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1 1
Uji Kompetensi 1.1
1
, ,
2 8 16 5
11
8
8
12
16 16 16
B. , 1,
A. 1. 54 x 20 – 640 : ( -8 ) = 1080 + 80
2
,
1
,
11 5
1, 12 , 8
= 1160
4
6
C. ,
2. 750 – 75 x ( -4 ) + 185 : 8
71
,
80
5 10 100
60
,
71
,
100 100 100
4 71 6 , , 5 100 10
= 750 + 300 + 23,125 = 1073,125
4 7 11
16 21 22
3 4
12 12 12
D. , ,
3. ( 300 + 1100 ) : ( -7 ) x 18 – 60
,
6
11 7 4
– 60
= 1400 : ( -7 ) x 18 = -200 x 18
– 60
= -3600
– 60
,
, ,
6
1
2. A. 5
= - 3660 4. 1.200 : ( -40 ) + 35 x 16 – 250
1
66
5
5
13 =
C.
1 =
D.
7 =
5. 8.000 : ( -50 ) – 2.800 + 60 x 9 = - 160 – 2.800 + 540
3
B.
= -30 + 560 – 250 = 280
16
=
3
4 3
7
15
8
8
3
52
7
7
= - 2420 3. A. B. 1. ( 48 – 2 ) x 21 x Rp 4.500,00 = 46 x 21 x Rp 4.500,00
B.
= Rp. 4.347.000,00 C.
2. 8 x 275 – 350 – 250 – 2 x 350 + 98 = 2.200 – 350 – 250 – 700 + 98
D.
= 998
3 2𝑎 16 3
+
−
14 10
+
8
=
2𝑎
11
3
3
3
−12
=
14
14
−10𝑎
𝑎
=
−6 7
−𝑎
=
10
4
=
2𝑎
5
− =
−9 9𝑎
5
10
3. 453.000 : 56 x 5 4. A.
= 8090 x 5
1 3
= 40.450
1
5
5
15
+ = 4
1
3
5
B. 1 + 3
4. ( 20 x 15 + 15 x 20 + 13 x 25 ) : 37 = ( 300 + 300 + 325 ) : 37 =
925
: 37 C.
5. Rp 100.000,00 x 5 – 75 x Rp 6.500,00 = Rp 500.000,00 – Rp 487.500
D.
5 6
1 1
1. A. , ,
1
2 8 16
8
,
2
,
1
8
24
Kunci jawaban Matematika kelas x
1
5
−
9 24
16
4
5
3
5
2𝑘
−
7 4𝑘
+
=5
1 6𝑘
=
48 15
11
=
7
5
15
+
15
1
15
35
8
3
83
15
=
=5
1 +3 = +
=
16 16 16
3
15
20
= E.
16
83
3
− =
= Rp 12.500,00
Uji Kompetensi 1.2
7
8
=
15
= +
=
= 25
3
+
24
=
35 15
+
48 15
8 15 30
12𝑘
−
21 12𝑘
+
2 12𝑘
11 12𝑘 Sagufindo Kinarya
F.
−6
+
6𝑎
5 −8𝑎
B. C. D.
E. F.
1 2 1 5 𝑥
1
6. A. B.
C.
D.
C.
51
5
81
𝑦
𝑧
𝑧
𝑥
31
42
x
1
:
3
21
105
5 10
3
11
4
2
4
4
2
3 3
3
3
1 : = ∶ 5
1
37
8
4
8
F.
7. A.
4𝑏 2 :
2𝑏
3𝑎2 2𝑎 5𝑏3
:
3𝑎2
=
7𝑏
5𝑏3
9
B.
=
4
34
x
82
2𝑏
2𝑎
7
=
11
C.
14
D.
=2 x
41
=
3
4
3
4
= =
4
11 10 4 77 6
(
3
4
+ x
1
1
6
3
3
1
7
1
4
9
4
(3 − 1) = × 2 7
9
7
3
9
4
4
4
11
3
42
+ )=
24
98 9
+
4
−1 19 62
+2
24
93
×
4 5
×
246 5
13
=
2
6
×
+
164
6
1539
5
5
12 10
+
=
5
4
=2( =2x
x
17
×
36 10
10
4
4
22
8
=2(
11
3
= 12
5
147
=
3
5
=9
24 9
884 150
+
21 24
=9
57 4
:
5
208 15
3
48 10
3 5
18 5
)
54
+
)
15
104
24 5
=4
4 5
1
3
3
5
=3 ×3 = =
102 3
×
18 51
36 3
= 12 cm2
15 13
= 13
15
L=pxl
1
3
4
4
=2(5 +2 ) 4
= 2 (7 4) 2
=
L=pxl
3
15
4
342
1
50
150
4
×
=
+
134
=5
153 34
B. k = 2 ( p x l )
6
Kunci jawaban Matematika kelas x
3
98
=2(3 +3 )
10𝑏2
3
3
8
9. A. k = 2 ( p + l )
21𝑎
4
11
9
=
= 18𝑏
3
x
7
17
34
1
10
1
1 (3 − 3 )
= +
3
=1 9
21
37
=
7𝑏
31
x
31
17
37
42
42
21
x
(3 x 2 ) + ( x 2 ) =
8
= 23
x5
513 905 5 = 392 + = = 25 36 36 36 36
11
=
2
∶
=2
9
= 4𝑏 2 x
9
185
7
=6
7
∶
1 2
25
=
1
1
4 :4 =
8. A.
102
3
2 :3 =
8
= × = =1
3
x
51
D.
6
3
x
3
=0
1
1
=
=1
51
2
3
37
6
81
x
x x3
5
8
3
(4 ) =
5
=
93
1
=1 x0
=1
51
x
51
37
=
= E.
24𝑎
=
3
2
B. 4 8 (1 3 + 3 3)
4
4
=
5
26
1
1
1
24𝑎
−
1𝑏 = 𝑏
x
2
24𝑎
−15
6
x1 x
𝑦
12𝑎
+
24𝑎
x1 =
8
−24
=
1
=
3
x
4
3
1
x
13
−65
= 5. A.
−
1
3
4
4
=5 ×2 =
21 4
×
11 4
Sagufindo Kinarya
7 8
231
=2x8
=
= 16 cm
= 14 16 cm2
76.400
x=
16
28.650
× 3 𝑑𝑜𝑙𝑎𝑟 = 7,9 𝑑𝑜𝑙𝑎𝑟
7
3
6. Timah putih 1
1
3
5
10. A. Matematika = 1 − − 15
=
15
7
5 15
−
= 37,5 kg 3
7. A. 1 cm
15
15
1 x=1:
15
=1×
× 21 orang
15 71
2100 𝑚 30 𝑚
8. A. 1 cm 1,5
30 m
cm
x=
: 4
x 4 3
1
× 30 𝑚 = 40𝑚
9. 4 cm
2. A. L1 : L2 = 6S12 : 6S22
6m
1,6 cm
= 6 . 32 : 6 . 42 =9
× 30 𝑚 = 45 𝑚
1
6
22
:
2100 m
B. 1 cm
= S12 : S22 12
30 m
1,5 cm
4
x=
: 16
B. V1 : V2 = S13 : S23 = 33 : 43
1,60,4 41
x
× 6 𝑚 = 2,4 𝑚
10. Ps = Pp : S
= 27 : 64 = 12 cm : 3. 300 ekor 240 ekor x=
300 240
× 1 𝑐𝑚 = 70 𝑐𝑚
× 213 orang
Uji Kompetensi 1. 3
=1
30 m 2100 m
x=
x
=
× 1 𝑐𝑚 = 20 𝑐𝑚
30 𝑚
x
= 45 orang
1. L1 : L2
6002 𝑚
B. 1 cm
x 7
600 m
x=
21 orang
12
30 m
x
7
= B.
−
= 4 × 50 kg
1 400
24 hari
400
= 12 cm ×
x
× 24 ℎ𝑎𝑟𝑖 = 30 ℎ𝑎𝑟𝑖
1
= 4800 cm = 48 m
4. 100 km
2 ,8 ℓ
x x=
2,814 0,2
5. 3 dolar x
ℓs = ℓp : S
0,2 ℓ
= 9 cm :
× 100 𝑘𝑚 = 1400 𝑘𝑚 = 9 cm ×
1 400 400
Rp 28.650,00
= 3600 cm
Rp 76.400,00
= 36 m
Kunci jawaban Matematika kelas x
3
1
Sagufindo Kinarya
11. A. B.
C. D.
3 51
Harga satuan
7 2
75
= 20.000 Laba
× 100% = 2%
100
× 100% =
68
=
12. A. 35 % =
35 100
12,5
B. 12,5 % =
=
C. 3%
B.
13 2 9
8
15
15.
9
115 100
=
23
=1
20
× 260 𝑘𝑚
Uji Kompetensi 1. 4
20
1. A. 34+7 = 311 B. (x + 1)8+6 = (x + 1)14
3
8
5
5
C. 2 (x3×4) = 2x12 𝑎
= 1,6
𝑎
D. ( 𝑏 )5+2 = ( 𝑏 )7 E. ( y3+2 )4 = ( y5 )4 = y5x4 = y20
= 0,73 75
242
90
804
3
= 1,54
25 6 × 100 % =
819
201
= 65 km
F. 14.
x
100
C. 1 = D.
x =
1
= 0,22
11
260 km
20 %
3
D. 115 % = 20
100
× 20.000 = 4000
3
20
=
100
% = 22 %
3
7
=
20
Harga jual = Rp 24.000,00
2
20. 80 %
13. A.
= Rp 2.800.000,00 : 140
× 1004 % = 28%
251
17
19. 7 kodi = 7 x 20 = 140 biji
× 10020 % = 60%
2
𝑎3𝑥4 𝑏 2𝑥4
1
𝑎4 𝑏 3
=
𝑎12 𝑏 8 𝑎4 𝑏 3
= a12-4 b8-3 = a8 b5
% = 37 % 2
2. A. B B. S
× 100% = 90%
C. S D. B
16. Potongan = Rp 11.500,00 – Rp 10.925,00
E. B
= Rp 575,00 % potongan =
575 11.500
F. B
× 100% = 5%
3. A. 6 x 103+5 = 6 x 108 B. 56 x 102 = 5,6 x 103
17. % untung = 100 % - 40 % - 15 % - 7,5 %
C. 960.000 = 9,6 x 105
= 37,5 % Untung =
37,5 100
D. 200 = 2 x 102
× 𝑅𝑝 2.750.000,00
E. 5 x 105-2 = 5 x 103 F. 18.600 = 1,86 x 104
= Rp 1.031.250,00 18. Bonus
1
4. A. ( 24 )4 = 2
15
= 100 × 𝑅𝑝 375.000 = 56.250
2
Gaji pokok
= 375.000
B. ( 33 )3 = 32 = 9
Gaji total
= 431.250
C. ( 82 )2 = 83 = 512
Kunci jawaban Matematika kelas x
3
4
Sagufindo Kinarya
1
1
3
4
F. X3 = 125
D. 4 x ( 43 ) - x ( 34 )
3
X = √125
= 4 x ( 4-1 ) x 3 =4x
=5
1
x3
4
=3 E. (
23
x
2
106
) = (( 2 x 3
102 )3)
6. A.
2 3
= ( 2 x 102 )2 = 4 x 104 1
F. 34 x ( 33 ) : 3 2
=3
B.
1 − 2
C.
3 1 4+2+2
E.
5. A. ( 25 )x = 2-4
1 2
1
𝑞
−3
𝑥
−2
3
𝑧
1 −1 2
=𝑧
−
1 2
7. A. √𝑎
5 x = -4
3
4
B. √𝑏
5
C. √𝑐 2
3
1
5
D. √𝑑 4
B. ( 3-1 )x -2 = 3-2 . 32 3-x+2 = 3
=𝑝
1 𝑝2
−
D. 𝑦 2
= 36
x=−
1
−1
1 2
8.
-x + 2 = −1
1
3 𝑥 103 𝑚 1 𝑑𝑒𝑡𝑖𝑘
=
3000 𝑚
5 = 𝑚⁄𝑗𝑎𝑚 3600 𝑗𝑎𝑚 6
2
9. 2,7 𝑥 10−23 𝑞 𝑥 4 𝑥 1028
1
32 = x
= 10,8 𝑥 10−23+28
C. x = 3 1
= 10,8 𝑥 105 𝑞
D. ( 52 )3x = 52 36x = 5 6x = x=
= 1,08 𝑥 106 𝑞
1 2
1
10. F =
2 1
𝑘 .𝑄.𝑞 𝑟2
𝑟2 =
12
𝑘 .𝑄 .𝑞 𝐹
4
E. ( 32 )x+3 = √(33 )𝑥 + 5 32x+6
=3
2x + 6 =
𝑟=√
3𝑥+15 4
3𝑥+15
= (𝑘
4
.
𝑘 .𝑄 .𝑞 𝐹
𝑄 .𝑞 1 𝐹
)2
8x + 24 = 3x + 15 8x – 3x = 15 – 24 5x = -9 x=−
9 5
Kunci jawaban Matematika kelas x
5
Sagufindo Kinarya
Uji Kompetensi 1. 5
F. 3√7 5. A. 2√15 + 2√10
1. A. Bukan B. Bukan
B. √48 - 2√18 = 4√3 - 6√2
C. Ya
C. (√3 )2 + 2 (√3 ) (√2 + (√2 )2
D. Bukan
= 3 + 2√6 + 4
E. Bukan
= 7 + 2√6
F. Ya
D. √6 x 6√6 = 6 x 6 = 36 E. 5.2 - √6 + 5√6 - 3
2. A. √27 = 3√3
= 10 + 4√6 - 3
B. √28 = 2√7
= 7 + 4√6
C. √25 .3 = 5√3
F. (√5 )2 - 2(√5 ) (√3 ) + (√3 )2
D. √49 .3 = 7√3
= 5 – 2√15 + 3
E. 2√49 .2 = 2.7√2 = 14√2
= 8 - 2√15
F. √25 .2 = 5√2 G. √100 .2 = 10√2
× √7 2 = √7 7 √7 × √7 √3 × √2 √6 B. = 2 √2 × √2 1 × 1−√3 1−√3 1−√3 C. = = 1+√3 × 1−√3 1−3 −2
6. A.
H. √100 .3 = 10√3 I. 5√4 .2 = 5.2√2 = 10√2 J. 10√100.10 = 10.10√10 = 100√10 3. A. c2 = a2 + b2 =
22
+
32
D.
2
2+√2 2−√2
=4 + 9
=
= 13 c = √13 (betuk akar)
× 2+√2 22+2 .2√2+(√2) = 4−2 × 2+√2
4+4√2+2 2
=
6+4√2 2
2
= 3 + 2√2
× √5 1 1 = √5 = × 2,236 = 0,45 5 5 √5 × √5 20 × √2 20 B. = √2 = 10 × 1,41 = 14,1 2 √2 × √2 10 × √6 10 5 C. = √6 = √2 . √3 6 3 √6 × √6
B. bentuk akar
7. A.
C. bukan D. bukan E. bentuk akar D. bukan
1
5
= 3 × 1,41 × 1,732 4. A. √9. 2 + 8 = 310√2 + 8
= 4,07
× √3 6 = √3 = 2 × 1,732 = 3,46 3 × √3 3 3 × √5 5 2,236 E. = √5 = = 1,12 2√5 × √5 10 2 4 × √2 4 4 F. = √2 = × 1,41 = 0,56 5√2 × √2 10 10
B. 5√3 - 2 √16.3 = 5√3 - 2.4√3
D.
= 5√3 - 8√3 = -3√3 C. 2√5 D. 15√13 + 9√13 - 2√3 + 3√3 = 24√3 - √3 2
6
3
E. √ 4 . 6 + √ 64 . 3 + 2√−81 3
= 2√6 + 4√ 3 + 3.9 3
= 2√6 + 4√ 3 + 27 Kunci jawaban Matematika kelas x
6
Sagufindo Kinarya
1
8. A. sekawan = √3 -1
B. x = 3-3 = 27
(√3 + 1) (√3 - 1) = (√3 )2 – 12 = 3 – 2 = 2
C. x4 = 16
B. sekawan = √5 +2
x=2 D. x-3 = 0,125
(√5 - 2) (√5 + 2) = (√5 )2 – 22 = 5 – 4 = 1 C. Sekawan = √7 + 3 (√7 - 3) (√7 + 3) = (√7 )2 – 32 = 7– 9 = - 2
1 𝑥3 1
D. Sekawan = 3 +√2 (3 -√2 ) (3 +√2 ) = 9.
125
x-3 =
𝑥
32
– (√2
)2
=9–2=7
1000 125
=
=
1000 5
10
x=2
L =pxℓ
5
3. A. log 2 × 4 = log 10 = 1
= (2+√2 ) (2-√2 ) B.
= 22 – (√2 )2
5log
155 ×25 62
=4–2
= 5log 125
= 2 cm2
=3
d2 = p2 + ℓ2
C. 3log 72 x 7log 33
= (2 + √2)2(2-√2 )2
2 . 3 x 3log 7 x 7log 3
= (4 + 4√2 +2) + (4 - 4√2 +2)
= 6 x 3log 3
= 12
=6 x 1 =6
d = √12 = 2√3 cm
4. A. 25 = x + 6 32 x + 6
10. AC = √22 + 12
26 x
= √4 + 1 B.
= √5
9
1 2
= 2x – 1
3 = 2x – 1
AD = √𝐴𝐶 2 + 𝐶𝐷2
4 = 2x
2
= √(√5) + 12
2=x
= √5 + 1
C.
= √6
10log
𝑥 16
103 =
5.
1. A. 5 B. 3log
1 − 3 3 .3 2
10log
D. 2log
10-3
3 −− 3 2 .=
= 3log
1 − 33 2
𝑥 16
=
A. log 24 = log (23 x 3 ) = log 23 + log 3
1 − 32
= 3 ( 0, 301 ) + 0, 477
= -3 -
=3
16.000 = x
Uji Kompetensi 1.6
C.
×2
= 0, 903 + 0, 477 = 1, 380
3 2
B. Log 600 = log ( 2 x 3 x 100 ) = log 2 + log 3 + log 100
2. A. x = 42 = 16
= 0, 301 + 0, 477 + 2 Kunci jawaban Matematika kelas x
7
Sagufindo Kinarya
= log 3,981 – log 104
= 2, 778 1
= 0,6 – 4
1
C. Log 2 . 32 = log 2 + log 32 1
= 0, 301 +
2
= -3,4
x 0, 477
= 0, 301 + 0, 2385
9. A. 0,3856
= 0, 5395
B. 3,477 C. -0,6198 1 2
D. -1,28735
D. Log 23 . 32 + log 3 . 5
1
= log 23 + log 32 + log 3 + log 52 10.
1
= 3(0,301)+2(0,477)+ 0,477 + (0,699) 2
6.
=
7.
C. 1479,1
= 2, 6835
D. 0,22
=
=
50 =
𝑙𝑜𝑔2 50
Uji Kompetensi 1.7
𝑙𝑜𝑔2 6
2 log 2 .52 2 log 2 .3
2𝑙𝑜𝑔2+2 log 52
=
1. ( 32 )2x – 1 =
2 log 2+2 log 3
34x-2
1+2 .𝐵
15 =
3 log 6 3 log 15
3 log 2 .3
3 log 2+3 log 3
𝑃+𝑄
35
= 3-5
4x
= -3
x
=
2. (25 )
= 3 log 3+2 log 3
3 log 3 .5 1+𝑃
1
4x – 2 = -5
1+𝐴
6log
=
B. 17,5
= 0, 903 + 1, 431 + 0, 3495
6log
=
A. 2,5
−3
2𝑥−3 2
𝑥+1 4
= (27 )
10𝑥 − 15 7𝑥 + 7 = 2 4
1 +1 𝑝
= ( )× 𝑃 1+𝑄
×𝑃
40– 60 = 14 + 14
1+𝑃
40–14 = 14 + 60
𝑃+𝑃𝑄
26 = 74
8. A. log 39, 81 = log ( 3, 981 x 10 ) = log 3, 981 – log 10
𝑥=
= 0, 6 – 1 = - 0, 4
(B)
4
74 26
=
37 13
(E)
4
B. Log 398100 = log 3, 981 x 105
3. 3
2𝑥−1 3
= log 3, 981 + log 105
2𝑥−1
= 0, 6 + 5
3
= 5,6
= 3−3
= −3
2 -1= -27
C. Log 0, 03981 = log 3, 981 : 102
2=-26
= log 3, 981 – log 102 = 0,6 – 2
= -13
= -1,4 D. Log 3,981 : 104 Kunci jawaban Matematika kelas x
8
Sagufindo Kinarya
2 4𝑥−2
4. ( ) 3
2 0
= 74 + 14√7 (E)
=( ) 3
4 -2 = 0
3
2
2
3
3
11. √54 + √16 − √250
1
= 4 = 4 = 2 (C)
3
3
3
= 3√2 + 2√2 − 5√2 = 0 (E)
5. P = 3 ℓ
d2 = p2 + ℓ2
12. 7log 23.2log32.3log 7-1 )2
L = 72
= ( 3√24
P x ℓ = 72
= 216 + 24
3ℓ x ℓ = 72
= 240
3ℓ2
+ 24
= 3.2(-1).7log2.2log 3.3log 7 = - 6. 7log 7 = - 6.1
d = √240
= 72
ℓ2 = 24
= -6
= 4√15
(C)
ℓ = √24 13.
p = 3ℓ
log 98 log 6
= 3√24 = 6.
r4
r3
x
=
r4+3-1
:
r1
= r6 = 2
4
=
log2 2 . 3
1+2 .2 log 7 1+𝑏
1+
=
log2 2 + log2 72 log2 2 + log2 3
2 log7 2
1+𝑏
2 𝑎
×𝑎 1+𝑏 ×𝑎 1+
=
log 2 .72
=2
𝑎+2
(C)
𝑎(𝑏+1)
1
7. (5−3 )−3 + (23 )3 − (103 )3 = 52 + 24 -10
14.
log5 75
=
log5 5
= 25 + 16 + 10
=p+2 4
=
9.
×1 − √5 4 ( 1− √5 ) = 12 −5 − √5
1+√5 ×1
4 ( 1−√5) −4
= −1 + √5
15. 3log 3
(A)
125 5
log 25
3
2−√3×2+√3
(D) 10
: 3log
2
log 25 : 3log 5
3
2 ×2+√3
=
1
= 5log 3 + 2 . 5log 5
= 31 (D)
8.
log5 3 . 52
= 3log 25 = 2 (A)
log 5
2(2 + √3) 2
22 − (√3)
16. 3log
41 . 54 27 82
= 3log 27 = 3 ( C )
= 4 + 2√3 (B) 17. 2log 5 x 5log 6 – 2log 24 = 2log 6 – 2log 24
10. (3√7 + 5)(4√7 − 2)
= 2log
= 12.7 − 6√7 + 20√7 − 10
2
24 1
= log
= 84 + 14√7 − 10 Kunci jawaban Matematika kelas x
6
4
9
Sagufindo Kinarya
= 14.000.000 cm
= -2 (A)
= 140 km
(B)
3
log 81
18. 15log 81 = = = =
4 3
log 3 × 5
3
log 15
=
9 3 35 25. 21 . - 2. + 42 2 8
4 3
log 3+ 3log 5
=
4
9 2
3
1+ log 5 4 1 1+ 𝑃
=4
×𝑃 4𝑃 = ×𝑃 𝑃+1
(B)
-
9
3
+4
2
8
3 8
26. ( E )
19. H Jual = 115.000 x 100 = 11.500.000 H Beli =
= 10.500.000 _
27. ( B )
Untung
= 1.000.000
28. L = 𝜋 r 2
1.000.000
% untung =
10.500.000
= 9,5 %
20.
V1
: V2
S13
: S23
23
r2 =
x 100 %
(E)
𝜋
r =√
. 24 . 24
1: 8
(D)
6010 366
= 3 √2 + 4√2 - 4√2 + 5√2 - 6 √2 = 2 √2
x 12
2
orang
=
2 10
= 24
= - 13
x 120 dm
(D)
33. P = 8 . ( 4-2 )1/2 . ( 64 )-1/3 = 8 . 4-1 . ( 43 )-1/3
x 120 dm
= 82 .
(C)
=
24. ∝s = ∝p : s = 7 cm :
( B )
32. 3 + 4√3 - 4 √3 - 16
= 20 orang ( E ) 2+3+5
( B )
31. 3 √2 + 2√4 . 2 - √16 . 2 + √25 . 2 – √36 . 2
(B)
36 hari x
2
( D)
𝑔
x 800 gr
250
22. 60 hari 12 orang
23.
𝐿
= 7 √15
= 1200 gr
x=
𝐿 𝑔
30. √735 = √15 .49
375 gr x 375
(D)
𝜋
= 2𝜋 √
21. 250 gr 800 gr
x=
𝐿
29. T = √4𝜋 2 .
: 43
12. 12. 12 : 2 4
𝐿
1
41
.
1 4
(D)
2
34. ( α2 + b2 )- ½ =
2.000.000
= 7 cm x 2.000.000 Kunci jawaban Matematika kelas x
1
1
= 10
1 √α2 + 𝑏2
1 (𝛼 2 + 𝑏2 )1/2
( C) Sagufindo Kinarya
35. 1 mol 1 2
6,02 . 1
mol
=
1023
x 60,2 . 1022
20
= 3,01 x 1022 7
24
36. (
)
3 .5
=
228 37 . 57
:(
312
x
212 . 53
x
=22 . 1013
23 . 3 52
9
5
. 3o
= 2.
(D)
3
44. 2. 4 − 3. 2 +
)
(D)
35 8
18 9 35 − + 4 2 8
510
36−36+35
215 . 35
8
= 228 – 12 15 . 312 – 7 – 5 . 510 – 3 – 7 = 21
. 26 . 108
27
= 23+6-7 . 105+8 ( E )
3
4
3 x( 4 ) 2 .5
105 . 23
=
35 8
3
= 4 8 (D)
45. E
. 5o Uji kompetensi 2.1
37. alog b . blog c . clog d = alog d
(A)
1. membilang = sesuatu yang eksak banyak siswa dalam suatu kelas
×3 − √5 4 ( 3− √5 ) 38. = 3+√5 ×3 − √5 9−5 4
)2
= ( 3 - √5
=
– 2 . 3√5 +
32
mengukur = pendekatan panjang , luas (√5)2
= 9 – 6√5 + 5
2. tiga cara :
= 14 – 6√5
1. pembulatan ke satuan terdekat
(D)
2. pembulatan ke angka desimal 39. log 23 . 3 . 53 = 2 log 2 + log 3 + 3 log 5 = 2 A + B + 3C
40. log ( x – 5 ) = log x-5 =
3. pembulatan ke banyaknya angka yang signifikan (berarti)
(D)
𝑥
3. A. 10, 84
5
B. 5, 42
𝑥
C. 157, 22
5
D. 1072 , 12
5x – 25 = x 4x =
25 4
=6
41. V = a
4. A. 0,1246 B. 0,008543
1
C. 172, 5
(B)
4
D. 1,281 5. A. 3 angka signifikan
d = s√3 3
S3 = a
B. 4 angka signifikan
= √𝑎 . √3 6
3
= √27𝑎2
S = √𝑎
C. 6 angka signifikan
(C)
D. 5 angka signifikan 42.
100.000 2.500.000
x 100% = 4 %
(C)
6. A. 17 B. 171
43.
105 27
10 3
26 . 104
( ) .( 5
23
C. 1708
2
)
Kunci jawaban Matematika kelas x
D. 17085 11
Sagufindo Kinarya
14. A. SM = ½ x 0,01 = 0,005 ≈ 0,01 kg 7. A. 45,1 SR =
B. 503
0,05
5
=
7,25
725
C. 0,1 % kesalahan =
D. 0,6
1
=
145
1 145
x 100% = 0,69
B. SM = ½ x 1 = 0,50 kg 8. A. membilang jumlah murid kelas X-1
SR =
B. mengukur panjang meja memakai penggaris
0,50 98
5
=
=
980
% kesalahan =
B. 10
SR =
C. 5,124 D. 199
0,5
1 196
100
5
=
1
1 200
200 x 100% = 0,50 %
D. SM = ½ x 0,1 = 0,05 menit
B. 1,71
SR =
0,05 9,5
5
=
950
=
C. 1,71 % kesalahan =
D. 1,714
11. Salah mutlak = ½ x satuan ukur terkecil Salah relatif =
𝑠𝑎𝑙𝑎ℎ 𝑚𝑢𝑡𝑙𝑎𝑘
SR =
ℎ𝑎𝑠𝑖𝑙 𝑝𝑒𝑛𝑔𝑢𝑘𝑢𝑟𝑎𝑛
1 km
749,5
B. 53,2 jam 0,1 jam 0,5 53,25
53,15
C. 8,01 detik 0,01
0,005 8,015
7,995
D. 45 cm
0,5
44,5
45,5
SR =
Salah relatif =
=
45,08
5 45080
=
8919
=
SR = 1
9016
25,10
=
89190
2510
=
1
37,05
=
x 100% = 0,53 %
1
=
154
1 154
x 100% = 0,65 %
0,05 3
=
5 30
1
= 1 6
6
x 100% = 16,67 %
0,005 90,05
=
5 90050
=
1 18010
1 18010
x 100% = 0,006 %
D. SM = 0,1 x ½ = 0,05 menit
17838
SR =
=
1 502
5 37050
Kunci jawaban Matematika kelas x
=
0,05 8,6
=
5 860
% kesalahan =
D. SM = 0,01 x ½ = 0,005 0,005
190
≈ 0,01 %
5
5
770
% kesalahan =
C. SM = ½ x 0,1 = 0,05 0,05
1
C. SM = 0,01 x ½ = 0,005 m
B. SM = ½ x 1 = 0,5 0,5
77
5
=
% kesalahan =
13. A. salah mutlak = ½ x 0,01 = 0,005 0,005
190
B. SM = ½ x 1 = 0,5 liter
0,5 750,5
1 cm
0,5
% kesalahan =
12. Pengukuran terkecil SM atas bawah A. 750 km
1
15. A. SM = ½ x 1 = 0,5 ton
persentase kesalahan = salah relatif x 100%
SR =
x 100% = 0,51
=
1000
% kesalahan =
10. A. 1,7
SR =
196
C. SM = ½ x 1 = 0,50 gr
9. A. 0,0018
SR =
1
=
1 172
1
172
x 100% = 0,58 %
1 7410
12
Sagufindo Kinarya
Uji kompetensi 2.2
SM = 0,005
C. pengukuran 1 = 5,51 gr
pengukuran 2 = 10,41 gr SM = 0,005 ------------------------------- + ----------------- + = 15,92 gr SM = 0,01
1. A. Pengukuran terbesar = 17,5 gr Pengukuran terkecil = 16,5 gr
Batas-batas pengukuran = 15,92 ± 0,01
B. pengukuran terbesar = 89,5 cm
Pengukuran terbesar =15,92+0,01=15,93
pengukuran terkecil = 88,5 cm
Pengukuran terkecil = `5,92–0,01=15,91
C. Pengukuran terbesar = 90,65 detik Pengukuran terkecil = 90,55 detik
7. A. pengukuran 1 = 15 kg
SM = 0,5 kg
pengukuran 2 = 10 kg SM = 0,5 kg ------------------------------- - --------------- + 5 kg SM = 1 kg
D. Pengukuran terbesar = 48,485 gr Pengukuran terkecil = 48,475 gr
Batas-batas pengukuran = ( 5 ± 1) kg
2. A. 8,8 – 8,3 = 0,5 cm
Pengkuran terbesar = 5 + 1 = 6 kg
B. 9,90 – 9,87 = 0,03 gr
Pengukuran terkecil = 5 – 1 = 4 kg
C. 77,8 – 77,4 = 0,4 kg
B. pengukuran 1 = 8,5 cm SM = 0,05 cm
D. 7 – 5 = 2 km
pengukuran 2 = 5,4 cm SM = 0,05 cm ------------------------------- + ------------------ + 3,1 cm SM = 0,1 cm
3. A. ( 12,5 ± 4,5 ) m B. ( 45,5 ± 0,05 ) kg
Batas-batas pengukuran = (3,1 ± 0,1) cm
C. ( 8,676 ± 0,06 ) mm
Pengukuran terbesar = 3,1 + 0,1= 3,2 cm
D. ( 0,6 ± 0,05 ) cm
Pengukuran terkecil = 3,1 – 0,1 = 3 cm
4. SM = ½ x 1 cm = 0,5 cm
C. pengukuran 1 =20,50 gr SM =0,005 gr
batas atas
= 167 + 0,5 = 167,5 cm
batas bawah
= 167 – 0,5 = 166,5 cm
pengukuran 2 =17,59 gr SM =0,005 gr ------------------------------- + ----------------- + = 2,91 gr SM = 0,01 gr
5. A. pengukuran terbesar = 15,4 + 0,1 = 15,5
Batas-batas pegukuran = (2,91±0,01) gr
pengukuran terkecil = 15,4 - 0,1 = 15,3 8. Keliling maksimum yang mungkin
toleransi = 15,5 – 15,3
K = 2 ( 12,5 + 15,5 )
= 0,2
= 56 cm
B. Toleransi = 75,17 – 75,13 = 0,04
Keliling minimum yang mungkin
C. Toleransi = 103,9 – 103,7 = 0,2
K = 2 ( 11,5 + 14,5 ) 6. A. pengukuran 1 = 8 kg
SM = 0,5 kg
= 52 cm
SM = 0,5 kg + + 23 kg SM = 1 kg Batas pengukuran = ( 23 ± 1) kg pengukuran 2 = 15 kg
9. Luas maksimum yang mungkin L = 10,5 x 7,5 = 78,75 cm2
Pengkuran terbesar = 23 + 1 = 24 kg
Luas minimum yang mungkin
Pengukuran terkecil = 23 – 1 = 22 kg
L = 9,5 x 6,5
= 61,75 cm2
B. pengukuran 1 = 15,1 cm SM = 0,05 10. Keliling maksimum
pengukuran 2 = 22,2 cm SM = 0,05 + + 37,3 cm SM = 0,1
K = 4 x 5,1
Keliling minimum
Pengkuran terbesar = 27,3+0,1= 27,4 cm
K = 4 x 4,9
Pengukuran terkecil = 27,3 - 0,1=27,2 cm Kunci jawaban Matematika kelas x
= 20,4 m
= 19,6 m
Luas maksimum 13
Sagufindo Kinarya
= 26,01 m2
L = 5,1 x 5,1
15. √30,25 = 5,5
Luas minimum
√20,25 = 4,5
L = 4,9 x 4,9
= 24,01
m2
5,5 + 4,5 = 5(D) 2
Uji Kompetensi 2.3
16. pengukuran 1 = 10 cm
SM = 0,5 cm
pengukuran 2 = 6 cm
SM = 0,5 cm
------------------------------ 1. C
------------------- +
= 4 cm
2. D
SM = 1 cm
P. max sisa = 4 cm + 1cm = 5 cm
(E)
3. B 4. A
17. D atau E
5. SR =
6. % =
0,005 3,75
0,05 28,2
=
5 3750
=
1 750
(D)
19. SM = ½ x 0,1 = 0,05
(A)
20. pengukuran 1 = 5,5 m SM = 0,05 pengukuran 2 = 3,8 m SM = 0,05 ------------------------------ ------------------- + = 1,7 m SM = 0,1 m
x 100%
= 0,177%
18. T = 7,6 – 6,8 = 0,8 (C)
(C)
Selisih minimum = 1,7 – 0,1 = 1,6 m
(E)
7. D 8. Toleransi = 375,72 – 375,48 = 0,24 ( C )
21. 24
0,5
7
0,5
9. A
25 +
0,5 +
10. 8, 95
56
1,5
8,87 0,08
56 ± 1,5 ( A ) (C) 1
11. Salah Mutlak = 2 × 0,01 = 0,005 ( D ) 12. pengukuran 1 = 5,7 m
SM = 0,05 m
pengukuran 2 = 3,1 m SM = 0,05 m ------------------------------ ------------------- + = 2,6 m SM = 0,1 m
0,5
8
0,5
8
0,5
8
0,5
32
2,0
32 ± 2,0 ( C ) 23. 3,2
Batas pengukuran terbesar = 2,6 + 0,1 = 2,7 m
22. 8
(C)
0,5
5,4 +
0,5 +
8,6
0,1
4,3 13. 125
0,5
245
0,5
143 +
0,5 +
513
1,5
4,3 ± 0,1 ( D ) 24. L max = 17,5 x 13,5 = 236,25 Lmin = 16,5 x 12,5 = 206,25
(D)
25. 25,5 x 15,5 = 395,25 ( D )
513 – 1,5 = 511,5 ( A ) 14. 8,45 x 6,45 = 54, 5025 ( B )
Kunci jawaban Matematika kelas x
14
Sagufindo Kinarya
1
Uji kompetansi 3.1
< 𝑥
2
∴ HP = { 1,2,3,… }
= 5𝑥 – 3
1. A. 4 + 10
=5𝑥 -4𝑥
10 + 3
3. A. 3 𝑥 + 2 𝑥 ≤ 7 𝑥 + 10
13 = 𝑥
3 – 5 + 6 ≤ 7 𝑥 + 10
∴ HP = { 13 } B. 4 + 4 𝑥
– 10 ≤ 7 𝑥 – 5 𝑥
=2𝑥 -6
– 10 ≤ 2 𝑥
4𝑥-2𝑥 = -6–4
– 10 ≤ 𝑥
2 𝑥 = - 10
∴ HP = { - 4 , - 3 , - 2 ,… }
𝑥 =-5 B.
∴ HP = { - 5 } C. 6 𝑥 - 3 + 1= 4 𝑥 – 2
1−2𝑥
−3<0
2−𝑥 1−2𝑥
6𝑥-4𝑥 =-2+3–1
−
2−𝑥
3(2−𝑥)
1−2𝑥−6+3𝑥
2𝑥 =0
2−𝑥
𝑥 =0
−5+𝑥
∴ HP = { 0 }
2−𝑥
D. 5 (2 𝑥 - 1) = 2 (𝑥 + 1) 10 𝑥 - 5
=2𝑥 +2
10 𝑥 - 2
=2+5
<0
<0
( -5 + x ) ( 2 – x ) < 0 √
x=5
+
7
2
8
∴ HP = {
7 8
}
6 𝑥 + 72
= 3 𝑥 – 27
6𝑥 -3𝑥
= – 27 – 72
3𝑥
= – 99
𝑥
= – 33
C.
4. A.
2𝑥−8
2 3
9𝑥+9−4𝑥+16
6𝑥−24
1
∈𝑅}
− >0
3(2𝑥−8)
>0
>0
(5𝑥 + 25)(6𝑥 − 24) > 0
2
x = -5
∴ HP = { 6,7,8,… }
+
B. 3 + 10 𝑥 -5 < 18 𝑥 – 6
8
3𝑥+3
,𝑥
3(3𝑥+3)−2(2𝑥−8)
2 𝑥 > 11
4
< x
Hp = { | 𝑥 > 3
3𝑥–𝑥 >5+6
4
<0
4
3
2. A. 3 𝑥 - 6 > 𝑥 + 5
3–5+6
3−𝑥
3–x<0
= { – 33 }
𝑥 > 5
5
Hp = { | 𝑥 < 2 atau 𝑥 > 5 }
E. 6 (𝑥 + 12) = 3 (𝑥 - 9)
∴ HP
x=2
-
8𝑥 =7 𝑥 =
<0
2−𝑥
-5
< 18 𝑥 – 10 𝑥
x=4 4
Hp = { | 𝑥 < -5 atau 𝑥 > 4 , 𝑥 ∈ 𝑅 }
< 8𝑥 < 𝑥
Kunci jawaban Matematika kelas x
15
Sagufindo Kinarya
B. 3 (7 - 4𝑥) ≤ 𝑥 + 1
HP = {
21 – 12𝑥 ≤ 𝑥 + 1 21 – 1 20
,1 }
( 2x – 1 ) ( x – 1 ) = 0
≤ 13𝑥
x=1 √ x=2
20 ≤ 𝑥 13 7 1 13 ≤ 𝑥
HP = { 1, 2 } G.
7 ∴ HP = { | 𝑥 ≥ 1 13
,𝑥
∈𝑅}
4+𝑥
=
4𝑥
3 𝑥+2
4x + 8 + x2 + 2x = 12x X2 + 2x + 4x – 12x + 8 = 0
C. 5 ( 2 𝑥 – 1 ) ≥ 3 ( 2 - 3 𝑥 )
X2 – 6x
≥ 6–9𝑥
+8=0
(x–4)(x–2)=0
10 𝑥 + 9𝑥 ≥ 6 + 5
x=4√ x=2
19 𝑥 ≥ 11
HP = { 2 , 4 }
11
𝑥 ≥
2
F. X2 – 2x + 1 – x + 1 = 0
≤ 𝑥 + 12 𝑥
10 𝑥 - 5
1
19
∴ HP = { | 𝑥 ≥
11 19
2. A. x2 – 4x
,𝑥
∈𝑅}
=0
X2 – 4x + 4 = 4 ( x – 2 )2
Uji Kompetensi 3.2 1. A. ( 𝑥 - 3 ) ( 𝑥 - 10 ) = 0
=4
x–2
= √4 = ±2
x1
=2+2=4
x2
= -2 + 2 = 0
HP = { 0 , 4 }
𝑥 = 3 √ 𝑥 = 10 ∴ HP = { 3 , 10 }
B. x2 + 3x
B. 3 𝑥 ( 3 – 𝑥 ) = 0
= 10 9
𝑥 =0√ 3=𝑥
(x+
∴ HP = { 0 , 3 } C. ( 𝑥 - 5 ) ( 𝑥 + 5 ) = 0
x+
3 2
)2 =
3 2
∴ HP = { - 5 , 5 }
x1
=
x2
=
D. 0 = 2 𝑥2 - 15𝑥 + 28 0 =(2𝑥 -7)(𝑥 -4) 2𝑥 = 7 √ 𝑥 = 4
49 4
=±
𝑥 =5 √ 𝑥 =-5
𝑥=
9
X2 + 3x + 4 = 10 + 4
7 2
7 2
3
4
2
2
− = =2 7
3
−10
2
2
2
− − =
= −5
HP = { -5 , 2 }
7 2
∴ HP = { 3
1 2
C. x2 – 2x ,4}
=3
X2 – 2x + 1= 3 + 1 ( x – 1 )2 = 4 x–1
=±2
( 2x – 1 )( x – 1 ) = 0
x1
=2+1=3
2x = 1 √ x = 1
x2
= -2 + 1 = -1
E. 2 x2 – 3x + 1 – x + 1 = 0
x =
HP = { -1 , 3 }
1 2
Kunci jawaban Matematika kelas x
16
Sagufindo Kinarya
D. X2 + 1 = 1
3. A. a = 1 , b = 7 , c = 3
X2 = 1 -1
−𝑏±√𝑏 2 −4𝑎𝑐
X1,2 =
X2 = 0
2𝑎
X = ±√0 =
X1 = 0 X2 = 0
= E.
2𝑥 2 +6𝑥=1
X2 + 3x + (x+
9 4
3 2 ) 2
=
x2
=
HP = { −
2
4
7 = 𝑥 2 − 2𝑥 2
4 √11 2
√11 2
3
− = 2
−√11 2
,
√11−3 2
3
−√11−3
2
2
− = √11−3 2
15+√305 8
X2 =
15−√305 8
}
C. a = 2 , b = 1 , c = -6
HP = {
G.
𝑥 2 +2𝑥=3
2 .2
= =
𝑥2 = =
−3√2 2
X2 =
−3√2+2 2
−1±7 4 4
−1−7
6
3
4
2
= =
=
4
−8 4
= −2
D. 7x2 + 13x – 2 = 0 −13±√132 −4 .7 .(−2)
x1,2 =
2 .7
:2 =
2
𝑥 + 2𝑥 + 1 = 3 + 1 ( x + 1 ) = ±2
=
X1 = 2 – 1 = 1 X2 = -2 – 1 = -3
X1 =
HP = { -3, 1 } X2 = Kunci jawaban Matematika kelas x
−1±√49 4
−1±7
X1 =
+1
−3√2+2 3√2+2 , } 2 2
2𝑥 2 +4𝑥=6
−1±√12 −4 .2 .(−6)
x1,2 =
3
= 𝑥1
15±√305 8
X1 =
± √2 = 𝑥 − 1 2
3√2+2 2
2 .4
=
:2
+ 1 = 𝑥1
−(−15)±√(−15)2 −4 .4 .(−5)
X1,2 =
= (𝑥 − 1)2
3√2 2
2
4x2 – 15x – 5 = 0
+ 1 = 𝑥 − 2𝑥 + 1
4
−7−√37
B. 4x2 = 15x += 5
2
18
−7±√37 2 2
X2 =
11
√11−3 2
7=2𝑥 2 −4𝑥
2
9
=±
2
x1
7
1
2 .1
−7±√37
X1 =
= +
=
3
x+
F.
:2
1 𝑥 2 +3𝑥= 2
−7±√72 −4 .1 .3
17
−13±√225 14 −13±15 14
−13±15 14 −13−15 14
= =
2 14
=
−28 14
1 7
= −2 Sagufindo Kinarya
E. a = 1 , b = -2q , c = -q2
= = =
X2 =
−𝑝−√𝑝 𝑝−1
−(−2𝑞)±√(−2𝑞)2 −4 .1 .(−𝑞) 2 .1
4. ( 2x – 2 )2 = x2 + ( x + 2 )2
2𝑞±√4𝑞2 +4𝑞
4x2 – 8x + 4 = x2 + x2 + 4x + 4
2
4x2 – x2 – x2 – 8x – 4x + 4 – 4 = 0
2𝑞±2√𝑞2 −𝑞
2x2 – 12x = 0
2
2x ( x – 6 ) = 0
= 𝑞 ± √𝑞 2 + 𝑞
x=0
X1 = q + √𝑞 2 + 𝑞 √𝑞 2
X2 = q -
√ x=6
sisi 1 = x = 6
+𝑞
sisi 2 = x + 2 = 6 + 2 = 8 sisi 3 = 2x – 2 = 2 . 6 – 2 = 10
F.
2𝑥−1+1(𝑥+1)
=
𝑥+1 2𝑥−1+𝑥+1 3𝑥
=
𝑥+1
5. A. a = 2 , b = 1 , c = 0
4
D = b2 – 4ac
𝑥
=
𝑥+1
𝑥
4
= 12 – 4 . 2 . 0
𝑥
= 1 > 0 memiliki 2 akar nyata yang
4
berlainan.
X2 + x = 12x
B. a = 1 , b = 2 , c = -3
X2 + x – 12x = 0 X2
D = b2 – 4ac
– 11x = 0
= 22 – 4 . 1 (-3)
a = 1 , b = -11 , c = 0 x1,2 = = X1 = X2 =
= 4 + 12
−(−11)±√(−1)2 −4 .1 .0
= 16 > 0 memiliki 2 akar nyata
2 .1
yang berlainan .
11+11
C. a = 4 , b = -12 , c = 9
2
11+11 2 11−11 2
=
22 2
D = ( -12 )2 – 4 . 4 . 9
= 11
= 144 – 144
0
= 0 = 0 memiliki 2 akar nyata yang
= =0 2
sama D. a = 2 , b = -3 , c = 4
G. a = ( p – 1 )
D = b2 – 4ac
b = 2p
= (-3)2 – 4 . 2 . 4
c=p x1,2 =
=
= = X1 =
= 9 – 32
−2𝑝±√(2𝑝)2 −4(𝑝−1)𝑝
= -23 < 0 tidak memiliki akar nyata
2 .(𝑝−1) −2𝑝±√4𝑝2 −4𝑝2 +4𝑝
6. A. a = 1 , b = m + 1 , c = 9
2 .(𝑝−1)
D=0
−𝑝±2√𝑝
b2 – 4ac = 0
2 .(𝑝−1)
( m + 1 )2 – 4 . 1 . 9 = 0
−𝑝±√𝑝
M2 + 2m + 1 – 36 = 0
𝑝−1
(m+7)(m–5)=0 m = -7 √ m = 5
−𝑝+√𝑝 𝑝−1
Kunci jawaban Matematika kelas x
18
Sagufindo Kinarya
B. a = m – 1 , b = 4 , c = -5
= -11 < 0 sehingga tidak memiliki akar
D=0
nyata ( imajiner )
b2 – 4ac = 0 42 – 4( m – 1 ).(-5) = 0 16 + 20m – 20 = 0
Uji Kompetensi 3.3
20 m = 0 4
m=
20
=
1
1. a = 1 , b = -3 , c = 2
5
−𝑏
A. 𝛼 + 𝛽 = 7. a = p , b = 2 , c = p b2 – 4ac > 0
C.
22 – 4 . p . p > 0 4–
𝑐
B. 𝛼 . 𝛽 =
D>0
4p2
2 1
𝛼+ 𝛽
=
𝛽
=3
= =2
𝑎
1
1
=
𝛼 .𝛽
3 2
= 32 − 2 . 2
( 2p + 2p )(2 – 2p) > 0 √
𝛼
+
−(−3)
D. 𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
>0
P = -1
1
𝑎
=
=9−4
p=1
=5 E. 𝛼 2 𝛽 + 𝛼𝛽 2 = 𝛼𝛽(𝛼 + 𝛽)
8. a = 1 , b = 2p – 1 , c = p2 – 2p + 3
= 2 .3
D<0
=6
– 4ac < 0 ( 2p –1)2 – 4 .1(p2 – 2p + 3) < 0
F.
4p2 – 4p + 1 – 4p2 + 8p – 12 < 0 4p – 11 < 0 4p p
<
𝛽
=
< 11
11
+
𝛽 𝛼
𝛼 2 +𝛽 2
=
𝛼𝛽
5 2
2. a = 3 , b = 5 , c = -2
4
<2
p
𝛼
𝛼+ 𝛽=
D = b2 – 4ac
3 4
𝛼. 𝛽=
= 52 – 4 . 3(-2) 9. a = p , b = - (2p – 3) , c = p + n
+ 4pn = 0 B.
-12p – 4pn + 9 = 0
2 𝛼
+
2 𝛽
=
9 = p ( 12 + 4n ) 12+4𝑛
=p
D=
3
– 4ac
= 32 – 4 . 1 . 5
√49 3
=
7 3
2(𝛼+𝛽) 𝛼𝛽
=
=
=
𝛼𝛽
2 .−
10. a = 1 , b = 3 , c = 5 b2
−2
2𝛼+2𝛽
=
9 = 12p + 4pn 9
√𝐷 𝑎
A. 𝛼 − 𝛽 =
( -(2p – 3))2 – 4 . p ( p + n ) = 0 – 12p + 9 –
=
= 49
b2 – 4ac = 0 4p2
𝑎
𝑎
=
= 25 + 24
D=0
4p2
𝑐
−𝑏
−2 3
−10 3
5 3
×−
3 2
=5
= 9 – 20 Kunci jawaban Matematika kelas x
19
Sagufindo Kinarya
−5 3
3. X2 + 3x – m = 0
1 1 2 12 12𝑥 2 +𝑥−1=0
𝑥 2 + 𝑥− =0
a = 1 , b = 3 , c = -m −𝑏
𝛼+𝛽 =
𝑎
=
−3 1
= −3
× 12
D. 𝛼 + 𝛽 = 2 − √3 + 2 + √3 = 4
𝛼 + 3𝛽 = 5
𝛼 . 𝛽 = (2 − √3)(2 + √3)
𝛼 + 𝛽 = −3
=4−3
2𝛽 = 8
=1
𝛽=4
𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼 . 𝛽 = 0
𝛼 + 𝛽 = −3
𝑥 2 − 4𝑥 + 1 = 0
𝛼 + 4 = −3 𝛼 = −7
-7 . 4 =
3
𝑎
1
7
3
3
=2 =
1
2
3
3
𝛼 .𝛽 = 2 . =
𝑐
𝛼 .𝛽 =
1
E. 𝛼 + 𝛽 = 2 +
𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼 . 𝛽 = 0
−𝑚 1
7
2
𝑥 2 + 3𝑥+ 3= 0
-28 = -m
3𝑥 2 − 7𝑥+2= 0
28 = m
×3
6. A. a = 1 , b = -4 , c = 2 4.
X2
– 6x +2p – 1 = 0
𝛼+𝛽 =
a = 1 , b = -6 , c = 2p – 1 D = b2 – 4ac
𝛼 .𝛽 =
= (-6)2 – 4 . 1 (2p – 1)2
𝑎
−(−4) 1
=4
2
= =2 1
X1 . x2 = 3𝛼 .3𝛽 = 9𝛼𝛽 = 9 .2 = 18
= 40 – 8p
4=
𝑐
𝑎
=
X1 + x2 = 3𝛼 + 3𝛽 = 3(𝛼 + 𝛽) = 3 .4 = 12
= 36 – 8p + 4
X2 – ( x1 + x2 ) x + x1 . x2 = 0
√𝐷 𝑎
𝛼− 𝛽=
−𝑏
X2 – 12
x + 18 = 0
B. X1 + x2 = 3𝛼 − 1 + 𝛽 − 1 = 𝛼 + 𝛽 − 2 =
√40−8𝑝
4−2= 2
1
x1. x2 =3(𝛼 − 1)(𝛽 − 1) = 𝛼𝛽 − 𝛼 − 𝛽 + 1
4 = √40 − 8𝑝
= 𝛼𝛽 − (𝛼 + 𝛽) + 1
16 = 40 – 8p
=2–4+1
8p = 40 – 16
= -1
8p = 24
x2 – ( x1 + x2 ) x + x1 . x2 = 0
P=3
x2 – 2 – 1 = 0
5. A. ( x – 2 )(x – 5 ) = 0 7. a = 1 , b =-2 , c = -3
X2 – 5x – 2x + 10 = 0
𝑥1 + 𝑥2 =
X2 – 7x + 10 = 0 B. ( x + 3 )(x – 7 ) = 0 X2
X1 . x2 =
– 7x + 3x – 21 = 0
𝑐 𝑎
=
X2 – 4x – 21 = 0 1
𝛼+𝛽 =
1
C. ( x – 4 )(x – 3 ) = 0 1
1
−𝑏 𝑎 −3 1
𝑥1 𝑥2
=
−(−2) 1
=2
= −3 𝑥2
+𝑥 = 1
𝑥12 +𝑥
22
𝑥1 𝑥2
1
X2 + 3x – 4x + 12 = 0 Kunci jawaban Matematika kelas x
20
Sagufindo Kinarya
=
= = =
(𝑥1 +𝑥2 )2 −2𝑥1 𝑥2
𝑥 2 − (𝑥1 + 𝑥2 )𝑥 + 𝑥1 . 𝑥2 = 0
𝑥1 𝑥2
7 𝑥 2 − (− ) 𝑥 + (−1) = 0 2 7 𝑥2 + 2 𝑥 − 1 =0 ×2 2𝑥 2 + 7𝑥 − 2 =0
22 −2(−3) (−3) 4 +6
c. 𝑥1 + 𝑥2 = 𝑥 2 + 𝛽+∝ + 𝛽 2
−3 −10 3
= −3
1
=∝3 + 𝛽 3
3
= (∝ +𝛽)3 − 3 ∝ 𝛽(∝ +𝛽) 1 3
1
= (2) − 3. (−2) (2)
2
8. 2𝑥 − 𝑥-4= 0
1 +3 8 25 = 8
a = 2, b = -1, c = -4
=
−𝑏 1 = 𝑎 2 𝑐 −4 ∝. 𝛽 = = = −2 𝑎 2 ∝ +𝛽 =
𝑥1 . 𝑥2 = (∝2 + 𝛽)(∝ +𝛽 2 ) =∝3 +∝2 𝛽 2 +∝ 𝛽 + 𝛽 3
a. 𝑥1 =∝ +1 𝑑𝑎𝑛 𝑥2 = 𝛽 + 1
=∝3 + 𝛽 3 + (∝ 𝛽)2 +∝ 𝛽
𝑥1 + 𝑥2 =∝ +1 + 𝛽 + 1
25 + (−2)2 + (−2) 8 25 = +4−2 8 25 = +2 8 41 = 8
=∝+𝛽+2
=
1
=2+ 2 5
=2 𝑥1 . 𝑥2 = (∝ +1)(𝛽 + 1) =∝ 𝛽+∝ + 𝛽 + 1 = −2 + =−
1 + 1 2
𝑥 2 − (𝑥1 + 𝑥2 )𝑥 + 𝑥1 . 𝑥2 = 0
1 2
25 41 𝑥2 − 8 𝑥 + 8 = 0 ×8 8𝑥 2 − 25𝑥 + 41 = 0
𝑥 2 − (𝑥1 + 𝑥2 ) + 𝑥1 . 𝑥2 = 0 5 1 2 2 2𝑥 2 −5𝑥−1=0
𝑥 2 − 𝑥+(− )=0
×2
9. X2 – x + 2 = 3a + 1 X2 – x – 3a + 2 – 1 = 0 X2 – x – 3a + 1 = 0
b. 𝑥1 =∝ −2 𝑑𝑎𝑛 𝑥2 = 𝛽 − 2
a = 1 , b = -1 , c = -3a + 1
𝑥1 + 𝑥2 = 𝛼 − 2 + 𝛽 − 2 = ∝ +𝛽 − 2
berkebalikan
1 −4 2 7 =− 2 𝑥1 . 𝑥2 = (∝ −2)(𝛽 − 2) =
x1 . x2 = 1 𝑐 𝑎
=1
−3𝑎+1 1
=1
−3𝑎 + 1 = 1
=∝ 𝛽 − 2 ∝ −2𝛽 + 4
1 − 1 = 3𝑎
=∝ 𝛽 − 2(∝ +𝛽) + 4
0 = 3a
1 = (−2) − 2. + 4 2
0=a
= −4 -1 + 4 = −1 Kunci jawaban Matematika kelas x
21
Sagufindo Kinarya
10. a = p – 2
5. x2 – 2x – 8 < 0 ( x – 4 )( x + 2 ) < 0
b = -2 c = 2 + 2a
x=4
misal
x1 = a
√
x = -2
+
0
x2 = 2b ( x – x1)(x – x2 ) = 0
+
√
4
HP = { x | -2 < x < 4 , x ∈ R }
( x – b )( x – 2b ) = 0 6. 3x2 – x2 – 9x + 4 ≤ 0
X2 – 2bx – bx + 2b2 = 0
2x2 – 9x + 4 ≤ 0
X2 – 3bx + 2b2 = 0
( 2x – 1 ) ( x – 4 ) ≤ 0 2x = 1
1.
x=4
1
x=2
Uji Kompetensi 3.4 1 2 𝑥 −2𝑥>0 2 𝑥 2 − 4𝑥>0
√
+
×2
+
1
√
4
2
1
HP = { x | 2 ≤ x ≤ 4 , x ∈ R }
𝑥 (𝑥 − 4) > 0 √
x=0
-
x=4
+
0
√
+
7.
4
3𝑥−2 𝑥
−3<0
3𝑥−2−3𝑥
HP = { x | x < 0 atau x > 4 , x ∈ R }
𝑥 −2
2. ( 2x + 3 ) ( x – 2 ) ≤ 0
𝑥
<0
<0
2x = -3 √ x = 2 x=
−3
+
2
+
−3
√
2
HP = { x |
−3 2
0
+ HP = { x | x > 0 , x ∈ R }
2 8.
≤ 𝑥 0 atau x ≤ 2 , x ∈ R }
𝑥−1 𝑥−2
−
𝑥−3 𝑥−4
≥0
(𝑥−1)(𝑥−4)−(𝑥−3)(𝑥−2)
3.
– 2x – 15 ≥ 0
(𝑥−2)(𝑥−4)
( x – 5 )( x + 3 ) ≥ 0
(𝑥 2 −5𝑥+4)−(𝑥 2 −5𝑥+6)
X2
x=5
√
+
x = -3 √
-3
(𝑥−2)(𝑥−4)
+
𝑥 2 −5𝑥+4−𝑥 2 +5𝑥−6 (𝑥−2)(𝑥−4)
5
HP = { x | x ≤ -3 atau x ≥ 5 , x ∈ R }
−2 (𝑥−2)(−4)
4. 6 + x – x2 < 0
-
√
≥0
≥0
-
x = -2
+ -2
≥0
𝑥 = 2,𝑥 −4
( 3 – x )( 2 + x ) < 0 x=3
≥0
√
√
-
+ 2
4
HP = { x | 2 < 𝑥 < 4 , x ∈ R }
3
HP = { x | x < -2 atau x > 3 , x ∈ R } Kunci jawaban Matematika kelas x
22
Sagufindo Kinarya
5=y L ≥ 136
9. K = 0
HP = { ( 2,5) }
2 ( p + l ) = 50
p x l ≥ 136
p + l = 25
p( 25 – p ) ≥ 136
l = 25 – p
25p – p2 ≥ 136 0≥
p2
3. 5x + y = 2
– 25p + 136
0 ≥ ( p – 8 )(p – 17 ) p = 8 √ p = 17
-
+ 8
5x + y = 5
17x + y = -8
5 (−
-12x
= 10
−
x
=−
-
10
=−
12
25 6
5
y
6
17
7x – 3y = 21
L 2 = s2
X2 + ( 576 – 48x + x2 ) > 20
11
2x2 – 48x + 576 – 20 > 0 2x2 – 48x + 556 > 0
3y =
X2 – 24x + 278 > 0 )
y= =
Uji Kompetensi 3.5
x + 2y = -2
x + 2y = -2
6
4 + 2y = -2
2x
=8
2y = -6
x
=4
y = -3
25
6 1 6
44x
= 84
x
= 44
84
21
11
−
147 11
84 11 11
:3
28
11
=2
6 11
5. 11x + 3y = -7
x5
55x + 15y = -35
2x + 5y = 21
x3
6x +15y = 63 49x x
= -98 = -2
2x + 5y = 21
HP = { ( 4,-3) }
2 (-2) + 5y = 21 2. 3x – y = 1
x2
x – 2y = -8
-4 + 5y = 21
6x – 2y = 2
5y = 21 + 4
x – 2y = -8 5x x
5y = 25
= 10
y=5
=2
HP = { ( -2,5 ) }
x – 2y = -8 2 – 2y = -8 2 + 8 = 2y 10 = 2y Kunci jawaban Matematika kelas x
23
10
= 11 = 1 11
231
84
6
55
35x +15y = 105
+ 3𝑦 = 21 3y =
+
x5
21
147
1. 3x + 2y = 6
30
=
9x – 15y = -21
7 . 11 + 3𝑦 = 21
L1 + L2 > 20
)(x
+y=5
x3
= (24 – x )2 = 576 – 48x + x2
(x
)+y=5
=9 4. 3x – 5y = -7
= x2
6
=
HP = { p | 0 < 𝑝 ≤ 8 atau 17 ≤ p < 25 }
10. L1 = s2
5
Sagufindo Kinarya
8. x – 4 = 0
6. 1 dan 2 2x + 5y + 3z = 6
2x + 5y + 3z= 6
6 x – 6y + 2z = 18 : 2 3x – 3y + z = 9
x=4 x2 – y2 = 16
x3
2x + 5y + 3z = 6
42 – y2 = 16
9x – 9y + 3z = 27
16 – 16 = y2
−7𝑥+14𝑦
=21
−𝑥+2𝑦
= −3
𝑥+𝑦
0 = y2
:7
0=y
=0
3𝑦
HP = { (4,0) }
= −3
y = -1
9. 4x2 – y = 0
x+y=0
4 . 22 – y = 0
x–1=0
4.4
=y
x =1
16 = y
2x + 5y + 3z
=6
HP = { (2,16) }
2 . 1 + 5 (-1) + 3z = 6 2–5
+ 3z = 6
10. x2 + y2 = 7
3z = 9
(√3 )2 + y2 = 7
z =3
3 + y2 = 7
HP = { (1,-1,3) }
Y2 = 4 y = ±√4
7. 1 dan 2 x+y+z=6
= ±2
x 3 3x + 3y + 3z= -3
x + 2y + 3z = 18
HP = { (√3, 2), (√3, −2) }
x + 2y + 3z = -2 2x + y
= -1
1 dan 3 Uji Kompetensi 3.6
x + y + z = -1
1. 3x – 2x < -7
3x – 2y – z = 2 4x – y
x < -7
=1
(B)
3
2
2. 123 . 4 (8𝑥 − 20) + 3,12 ≤ 124 . 3 (6𝑥 + 15) −
4 dan 5 2x + y = -1
4.12
4x – y = 1
9(8𝑥 − 20) + 36
≤ 8(6𝑥 + 15) − 48
6x
=0
72𝑥 − 180 + 36
≤ 48𝑥 + 120 − 48
x
=0
72𝑥 − 48𝑥
≤ 120 = 48 + 180 − 36
2x + y = -1
24𝑥
2 . 0 + y = -1
≤ 216
𝑥
y = -1
3.
x + y + z = -1 0 + (-1) + z = -1
2−3𝑥 >4 5
2−3𝑥 >20
≤9(C)
×5
−18 > 3𝑥
z = -1 + 1
−6 > 𝑥 ( B )
=0 HP = { (0,-1,0) } Kunci jawaban Matematika kelas x
24
Sagufindo Kinarya
2
8. x1 + x2 = 1 – √3 + 1 + √3 = 2
4. 3. 3 (6𝑥 − 12) ≥ 3.2 (6𝑥 + 2) 12𝑥 − 24 ≥ 36𝑥 + 12
x1 . x2 = ( 1 – √3 )( 1 + √3) = 1 − 3 = −2
−24 − 12 ≥ 36𝑥 − 12𝑥
x2 – ( x1 + x2 ) x + x1 . x2 = 0
2
–
2x
2
=0 (B)
9. –
−363 ≥𝑥 242 −3
–
x2
−36 ≥ 24𝑥
3
10. 𝑥1 + 𝑥2 = 1 = 3
≥ 𝑥 (A) 𝑥1 . 𝑥2 =
5. 2 + 1 < 3𝑥 − 1 + 1 < 8 + 1
−4 = −4 1
𝑥12 + 𝑥22 = (𝑥1 + 𝑥2 )2 − 2𝑥1 𝑥2
3 < 3𝑥 < 9 :3 1<𝑥 <3
= 32 − 2(−4) =9+8
6. a = 1 , b = m +1 , c = 2m – 1
= 17
D<0
11. x2 + x – 2 ≥ 0
b2 – 4ac < 0 (m+
1)2
( x + 2 ) ( x -1 ) ≥ 0
– 4 .1( 2m – 1 ) < 0
m2
+ 2m + 1 – 8m + 4 < 0
m2
– 6m + 5 < 0
x = -2 √ x = 1 +
-
+
-2 √ 1
( m – 1 )( m – 5 ) < 0
X ≤ -2 atau x ≥ 1
(A)
m=1√m=5 12. 2c + 3k = 101.500 +
-
√
+
1
c + 2k = 53.500
5
2c + 4k = 107.000
HP = { m | 1 < m < 5 , m ∈ R }
2c + 3k = 101.500 k=
7. a = 2 , b = -1 , c = -5
𝑥1 + 𝑥2 = X1 . x2 =
𝑐 𝑎
=
−𝑏 𝑎
=
2
=
−(−1) 2
=
1
5.500
c + 11.000 = 53.500
2
c
−5
= 42.500
5.500 + 42.500 = 48.000 ( B )
2
+2=
-
c + 25.500 = 53.500
13. x – y = 5
𝛼 + 𝛽 = x1 + 1 + x2 + 1 = x1 + x2 + 2 1
x2
x
5
x=5+y
=5+y
=5+2
x2 – y2 = 45
2
=7
( 5 + y )2 – y2 = 45 25 + 10y + y2 – y2 = 45
𝛼 . 𝛽 = ( x1 + 1 )( x2 + 1 ) = x1x2 + x1 + x2 +1 =−
5 2
1
+ +1
10 y
= 45 – 25
10 y
= 20 y=2
2
HP = { ( 7,2 ) } ( A )
= -1 X2 – ( 𝛼 + 𝛽)𝑥 + 𝛼 . 𝛽 = 0 5
x2 – 2 + ( −1 ) 2𝑥 2 − 5𝑥 − 2 Kunci jawaban Matematika kelas x
=0 =0
×2
(D) 25
Sagufindo Kinarya
14. 4p + 3q =
853.000
x5
3x
= -9
3p + 5q = 1.022.000
x3
x
= -3
20 p + 15 q = 4.265.000
4x + 6y = 4 (-3) + 6(3)
9 p + 15q = 3.066.000
= -12 + 18
11 p
=6
= 1.199.000
p
=
109.000
(A) 18.
2𝑏+2𝑝=8.800 𝑏+𝑝=4.400
15. Ayah = x
:2
b = 600 + p
Anak = y
600 + p + p = 4.400
(x–2)=6(y–2) x–2
(E)
600 + 2p = 4.400
= 6y – 12
2p = 3.800
x = 6y – 10
p = 1900
( x + 18 ) = 2 ( y + 18 )
b = 600 + 1900
x + 18 = 2y + 36
= 2500 ( E )
x = 2y + 18 6y – 10 = 2y + 18 4y
19. 10x + 4y = 31.000
= 28
4x + 10y = 25.000
y
=7
x
= 6y – 10
20x + 8y = 62.000 20x + 50y= 125.000
= 6 . 7 – 10
-
-42y= - 63.000
= 42 – 10
y = 1500 ( A )
= 32
(B) 20. 3b + 2p = 10.500
16. 3x + 4y = 17
x 3 21x + 28y = 119
5x + 7y = 29
x 4 20x + 28y = 116 x
b = p + 1000 3 (p + 1000) + 2p = 10.500
=3
3p + 3000 + 2p = 10.500
3x + 4y = 17
5p
= 7.500
p
= 1.500
b
= 1.500 + 1.000
3 . 3 + 4y = 17 9 + 4y = 17 4y = 8
= 2.500 ( E )
y =2
21. 2x + y
HP = { ( 3,2 ) }
x + 3 z = 16
17. 2x - 5y = -21
2x + y
3x + 2y = -3
x2
=5
2x + 6z = 32
6x – 15y = -63 6x + 4y = - 6
=5
-
y – 6z = -27 -
5y - z
x5
= 10
-19y = -57 y=3
5y – 30z = -135
3x + 2.3 = -3 3x
5y - z
=-3–6
Kunci jawaban Matematika kelas x
26
=
10 -
29z = -145 Sagufindo Kinarya
26. x2 + 5x – m = 0
z=5 5y – 5 = 10
a = 9, b = 5, c = -m
5y
= 15
𝛼 .𝛽 = 3
y
=3
𝑐 𝑎
x + 3.5 = 16 x
=3
−𝑚 9
=1
=3
−𝑚 = 27
{(1, 3, 5)}
𝑚 = −27
(D)
27. akar – akar kembar
22. 6x2 – 11x – 10 < 0
D=0 b2 – 4ac = 0
( 2x – 5 ) ( 3x + 2 ) < 0 2x = 5 √ 3x = -2 x=
5
=0
144 – 16a
=0
−2
x=
2
( -12 )2 – 4.a.4
3
144 = 16a 9 =a
+
−2
√
3
{x|
−2 3
<x<
5
9x2 – 12x + 4 = 0
+ 5
( 3x – 2 )2
2
3x – 2 = 0 3x = 0
,x∈R}
2
=0
2
x=3
23. D
(C)
28. A
24. X2 + x – 12 < 0
29. 3x2 – 4x + 1 = 0
(x+4)(x–3)<0
( 3x – 1 ) ( x – 1 ) = 0
x = -4 √ x = 3
3x = 1 √ x= 1 +
-4
1
x=3
+ √
3
30. Akar khayal
(E) D<0 b2 – 4ac < 0
{ x | -4 < x < 3 , x ∈ R }
52 – 4(p – 2 ).(-2) < 0
{ -3, -2, -1, 0, 1, 2 }
25 + 8( p – 2 ) < 0
25. a = 1, b = 2 , c = -a
25 + 8p – 16 < 0
D=
b2
– 4ac
=
22
– 4 . 1 ( -a )
8p
< -9 p<
= 4 + 4a
−9 8
(A)
31. ( x – 7 )( x + 3 ) = 0
X1 – x2 = 8
X2 + 3x – 7x – 21 = 0
√𝐷 =8 𝑎
X2 – 4x – 21 = 0
√4+4𝑎 =8 1
32. ( x – √2 )( x – √8 ) = 0 X2 – √8 x – √2 x + √16 = 0
√4 + 4𝑎 = 8
X2 – 3 √2 x + 4 = 0
4 + 4𝑎 = 64
(E)
4a = 60 a = 15
(A)
Kunci jawaban Matematika kelas x
27
Sagufindo Kinarya
33. X2 + x – 6 = 0
A. Baris = 5
( x + 3 )( x – 2 ) = 0
Kolom = 4
X1 = -3 √ x2 = 2
B. 5, 7, 7, 5
𝛼 = -3 + 2 𝛽 = 2 + 2
C. 7, 7, 8, 6, 8
= -1
D. 4
=4
1 2 2. A = [5 6 9 1
(x+1)(x–4)=0 X2 – 3x – 4 = 0 ( A )
3 4 1 5 9 7 8] transposenya [2 6 1 ] 2 3 3 7 2 4
B = [1
2 3 4 5 6]
34. A 𝑐
Berkebalikan
=1
𝑎
−𝑏
35. Berlawanan
𝑎
=0 C = [ [1 2
36. a = b, b = c, c = -a x1,2 =
= =
−𝑏±√𝑏2 −4𝑎𝑐 2𝑎
1 2 3 4 5 6 1 2 3 4 5
−𝑐±√𝑐 2 −4𝑏(−𝑎) 2.𝑏
1 3
−𝑐±√𝑐 2 −4𝑎𝑏
(D)
2𝑏
D=
−𝑏
x1 + x2 = X1 . x2 =
𝑎 𝑐
=
𝑎
−𝑏
=
−𝑎
𝑐 −𝑎
=
=
−𝑐 𝑎
(E)
−𝑏
= 9x2 + 12x − 36
𝑎
=
∶3
3. – (A)
3 1 −2 4 6 ] + 3[ ]=[ −2 0 5 −3 −4 −6 12 [ ] 15 −9 0 14 =[ ] 11 −9 3 1 −2 4 3 B. ([ ]+[ ]) − ([ −2 0 5 −3 −2 2 1 [ ]) 1 0 1 5 5 2 =[ ]−[ ] 3 −3 −1 0 −4 3 = [ ] 4 −3 2 1 3 1 C. 3 [ ] −2 [ ] 1 0 −2 0
4. A. 2 [
3 1
1
(E)
40. E
Uji Kompetensi 4 . 1 5 7 7 5 6 8 7 3 8 7 8 2 6 4 6 1 7 6 8 0 Kunci jawaban Matematika kelas x
5 7 9 ] 6 8 1
7 2 4 3 3 −1 1. [ ]+[ ]=[ ] 4 −5 2 −1 2 −4 2𝑥 8 4𝑥 −10 6𝑥 −2 2. [ ]+[ ]=[ ] 3𝑥 −7 2𝑥 12 5𝑥 5
−1
( x1 + x2 )2 = ( − 3 )2 = 9
1 3 2 4
Uji Kompetensi 4 . 2
= 9x2 + x2 – x2 + 12x – 36
0 =3𝑥 2 +4𝑥−12
39. X1 + x2 =
[
𝑎
36 – 12x + x2 = 9x2 + x2 0
5 6
𝑏
38. ( 6 – x )2 = (3x)2 + x2
0
2 4
7 8 9 1
37. a = -a, b = b, c = c
1. P =
3 4 5]
8 3
28
2 ]+ 0
1 ]+ 0
Sagufindo Kinarya
6 3 0 = [ 7
3 6 2 ]− [ ] 0 −4 0 1 ] 0 4 2 −2 3. [−1] − [ 7 ] = [−8] 3 −6 9 4. –
2+0 6+0 2 0 1 3 ][ ]=[ ]= 1 + 6 3 + 12 1 3 2 4 2 6 [ ] 7 15 2 0 −2 4 BC= [ ][ ]= 1 3 1 5 −4 + 0 8 + 0 −4 8 [ ]= [ ] −2 + 3 4 + 15 1 19
4 −2 ] 3 1 4 2 B = [−3 −1] 0 −5 1 −7 2 −3 6. A. A = [ ]− [ ] 8 0 4 −1 2 −4 = [ ] 4 1 −2 −4 2 3 B. A = [ ]+ [ ] 7 5 1 4 0 −1 = [ ] 8 9
1 2 B. At = [ ] 3 4 2 1 Bt = [ ] 0 3 −2 1 Ct = [ ] 4 5 5 9 −2 4 C. ( AB )C = [ ][ ]= 8 12 1 5 −10 + 9 20 + 45 −1 65 [ ]= [ ] −16 + 12 32 + 60 −4 92 1 3 −4 8 D. A ( BC ) = [ ][ ]= 2 4 1 19 −4 + 3 8 + 57 −1 65 [ ]= [ ] −8 + 4 16 + 76 −4 92
= [
BA =[
5. A = [
7. 3 c = 4b C=
E. AB – BC = [5
4𝑏
8
−4 8 9 9 ]−[ ]=[ 1 19 7 12
1 ] −7
3
8. X + 2 – 4 = 3
F. A ( B – C ) = [
X–2=3 X
=3+2
7+y+1=y 9. 2x + 3 + 2 = 7 2x = 2 x =1 3+y+2=5 y =0 x+y=1+0=1 −10 3 −7 8 ]−[ ] 12 5 4 −6 −18 ] 18 −9 ] 9
I. AtBt = ( BA )t BtAt = ( AB )t 3 24 15 2. A. [ ] [8 5] = [ ] 4 32 20 4 3 2 −3 1 B. [ ] [−2 1] = 4 0 2 0 2 8+6+0 6−3+2 [ ] 16 + 0 + 0 12 + 0 + 4
Uji Kompetensi 4 . 3 1 3 2 ][ 2 4 1 5 9 [ ] 8 12
1. A. AB = [
2+3 0 ]=[ 4+4 13
Kunci jawaban Matematika kelas x
0 −2 4 ]−[ ]) 3 1 5
1 3 4 −4 =[ ][ ] 2 4 0 −2 4 + 0 −4 − 6 4 −10 =[ ]=[ ] 8 + 0 −8 − 8 8 −16 5 8 G. ( AB )t = [ ] 9 12 2+0 1+6 1 2 2 1 H. AtBt = ( )( )=[ ]= 6 + 0 3 + 12 3 4 0 3 2 7 [ ] 6 15 2+3 4+4 2 1 1 2 BtAt = [ ][ ]=[ ]= 0 + 9 0 + 12 0 3 3 4 5 8 [ ] 9 12
=5
9 3 10. 2p = [ 7 6 6 10 =[ 2 2 3 5 P=[ 1 1
1 3 2 ] ([ 2 4 1
0+9 ]= 0 + 12
29
Sagufindo Kinarya
B. D = 3 . 2 – 1 . 6 = 6 – 6 = 0
14 5 =[ ] 16 16 px + qy p q x 3. A = [ ] [ ] = [ rx + sy ] r s y 2 B= [ 5
3 −1 0 3 C. |1 2 3| 1 4 3 2 4
2x + 3y 3 x ] [y] = [ ] 5x + 7y 7
4. 2p – 6q = -16
D = ( 12 – 12 + 0 ) – ( 0 – 27 – 2 ) = 29
p – 3q = -8
x5
2 1 0 2 1 D. |−1 3 2| −1 3 4 2 5 4 2
-5p + 8q = -13 -5p + 8q = -13 3x + 3y + 3z= -3
D = ( 30 + 8 – 0 ) – ( 0 + 8 – 5 )
x + 2y + 3z = 18
= 38 – 3
-7q = -53 q=
= 35
53 7
p – 3q = -8 p–3. p-
159 7
53 7
2. A. = -8 −56
=
B.
7
p=
10 2.10−6.3 −3
0 2 ][ 1 5 3 1 ][ 7 0
3 2 ]=[ 7 5 0 2 ]=[ 1 5
3 ] 7 3 ] 7
C. D.
B. Ya E.
1 10 −6 ]= [ 2 −3 2
[
1
−2 2(−2)−5(−1) −5
[
5
−6 ]= 2
−3
[− 3
1
2
1 1 −2 ] = −6+5 [ 3 −5
]
1 ]= 3
−1 ] −3
[
7
1 0 2 AI=[ 5
1
2 5
103
5. A. I A = [
−1 2 3
1
3 2.3−1.5 −1
1 3 −5 ] = 1[ 2 −1
[
1
3 [ 5.3−2.7 −7
1 3 −2 ] = 1[ 5 −7
1
3 [ 3.3−5.2 −2
−5 ]= 3
1 −1
[
−5 3 ]=[ 2 −1 −2 3 ]=[ 5 −7
3 −2
−5 ] 2 −2 ] 5
−5 −3 ]=[ 3 2
5 ] −3
3. A. x ( x – 5 ) – 2 . 3 = 0
6. A.
X2 – 5x – 6 = 0
B. Tidak
(x–6)(x+1)=0 x = 6 √ x = -1
1+3 1+2 1 1 1 1 ][ ]=[ ]= 3+6 3+4 3 2 3 2 4 3 [ ] 9 7 4+9 4+6 4 3 1 1 P3 = [ ][ ]=[ ]= 9 + 21 9 + 14 9 7 3 2 13 10 [ ] 30 23 −1 2 −1 2 8. A2 = [ ][ ]= 1 3 1 3 1 + 2 −2 + 6 3 4 [ ]=[ ] −1 + 3 2 + 9 2 11 7. P2 = [
1 3 −1 1 3 B. | 3 8 2 | 3 8 −1 x −3 −1 x ( -24 + ( -6 ) – 3x ) – ( 8 + 2x – 27 ) = 0 -30 – 3x + 19 – 2x -11 = 5x − 3.5−2.7
=
−1 2 4 ] − 2[ ]−5=0 1 3 11 −2 4 4 ]−[ ]−5 =0 2 6 11 0 ]−5 =0 5
[
[
5 −7
11 5
−2 ]= 3
=𝑥 1
15−14
5 −7
[
−2 ] 3
5 −2 ] −7 +3
B. AA1=[3 2] [−2 1] = [15 − 14 7 5 −5 3
1 =[ 0 C. A1A=[ 5
−7
−6 + 6 ] 35 − 35 −14 + 15
0 ] 1 −2 3 ][ 3 7
2 15 − 14 ]=[ 5 −21 + 21
10 − 10 ] −14 + 15
1 0 =[ ] 0 1
Uji Kompetensi 4 . 4
D. AA-1 = A-1A = I
1. A. D = 4 ( -2 ) – 3 ( -2 ) = -8 + 6 = -2 Kunci jawaban Matematika kelas x
1
4. A. A-1 =
A2 – 2A – 5 = 0 3 [ 2 3 [ 2 5 [ 0
=0
30
Sagufindo Kinarya
3 1 9 5. A. [ ]A = [ ] 3 2 12 A=
2 6−3 −3 1
[
3 −1 −2 4 ][ ] 6 0 −4 2 1 −6 − 6 12 − 0 x = 2[ ] 8 + 12 −16 + 0 1 −12 12 x = 2[ ] 20 −16 −6 6 x=[ ] 10 −8 −4 6 2 3 B. [ ]x = [ ] −7 8 1 4 1 4 −3 −4 6 x = 8−3 [ ][ ] −1 2 −7 8 1 −16 + 21 24 − 24 x = 5[ ] 4 − 14 −6 + 16 1 0 x=[ ] −2 2 5 2 6 5 C. x [ ]=[ ] 3 1 1 0 6 5 1 0 −5 x =[ ] [ ] 1 0 0−5 −1 6 0 − 5 −30 + 30 1 =[ ] 0−0 −5 + 0 −5 1 0 =[ ] 0 1 2 1 −2 4 D. x [ ]=[ ] 8 5 0 6 −2 4 1 5 −1 x =[ ] 10−8 [ ] 0 6 −8 2 −10 − 32 2 + 8 1 =[ ] 0 − 48 0 + 12 2 −21 5 =[ ] −24 6 1
x = 6−4 [
−1 9 ][ ] 3 12
18 − 12 ] −27 + 36 1 6 = 3[ ] 9 2 =[ ] 3 1 4 6 18 B. A – [ ]=[ ] 2 5 1 7 6 18 1 4 A=[ ]+[ ] 2 5 1 7 7 22 =[ ] 3 12 −8 2 −5 x 6. A. [ ] [y] = [ ] −13 3 −8 x 1 −8 5 −8 [y] = −16+15 [ ][ ] −3 2 −13 1 64 − 65 = −1 [ ] 24 − 26 x 1 [y] = [ ] 2 1 2 x 11 B. [ ] [y] = [ ] 2 −1 2 x 1 −1 −2 11 [y] = −1−4 [ ][ ] −2 1 2 1 −11 − 4 = −5 [ ] −22 + 2 1 −15 = −5 [ ] −20 x 3 [y] = [ ] 4 1 2 5 1 2 7. A. |1 x =0 5 |1 x 3 −1 −2 3 −1 1
=3[
9. – −2 1 0 −2 1 10. | 1 −2 1| 1 −2 2 −3 0 2 −3 D=(0+2–0)–(0+6+0)
( -2x + 30 – 5 ) – ( 15x – 5 – 4 ) = 0
=2–6
-2x + 25 – 15x + 9 = 0 34
= -4
=0
−2 1 ]=0+3=3 −3 0 1 1 A12 = [ ] = −(0 − 2) = 2 2 0 1 −2 A13 = [ ] = −3 + 4 = 1 2 −3 1 0 A21 = [ ] = −(0 + 0) = 0 −3 0 −2 0 A22 = [ ]=0+0=0 2 0 −2 1 A23 = [ ] = −(6 − 2) = −4 2 −3 1 0 A31 = [ ]=1+0=1 −2 1 A11 = [
2=x 1 2 B. |1 x 1 4
−3 1 2 −3| 1 x x 1 4
=0
( x2 – 6 – 12 ) – ( -3x – 12 + 2x ) = 0 X2 – 18 + 3x + 12 – 2x = 0 X2 + x – 6 = 0 (x+3)(x–2)=0 x = -3 √ x = 2 2 8. A. [ 4
1 −2 4 ]x = [ ] 3 6 0
Kunci jawaban Matematika kelas x
31
Sagufindo Kinarya
−2 0 ] = −(−2 − 0) = 2 1 1 −2 1 A33 = [ ]= 4−1= 3 1 −2 3 0 1 [2 0 2] 𝐴𝑑𝑗 𝐴 A-1 = = 1 −4 3 |𝐴| −4 A32 = [
− = −
3 4 1 2 1
0 − 0 −
7. (
1 5 5 −2 −2 5 −2 [ ] = 1[ ]=[ ] −7 3 −7 3 −7 3 1 4 −6 8 9. 𝑥 = 20−18 [ ][ ] −3 5 6 1 32 − 36 = 2[ ] −24 + 30 1 −4 = 2[ ] 6 −2 =[ ] (B) 3
8.
1 4 1 2 3
[− 4 1 − 4]
x = A-1B
1. a = 2
1 −2 −2 5 2 = 2−6 [ ][ ] −3 −1 −3 −6 1 −10 + 6 −4 + 12 = −4 [ ] −15 + 3 −6 + 6 1 −4 8 =−4 [ ] −12 0 1 −2 =[ ] (D) 3 0 0 1 0 1 11. A2 = [ ][ ] −2 3 −2 3 0−2 0+3 =[ ] 0 − 6 −2 + 9 −2 3 =[ ] (C) −6 7 2 3 1 2 3 12. |0 −1 2| 0 −1 2 4 1 2 4
b = 2a = 2 . 2 = 4 (D)
−2 3 −7 −5 1 2. A + B = [ ]+[ ]=[ −1 5 3 4 −3
4 ] (A) 2
3. –a = 4 a = -4 2a = 3a = 3 (-4) = -12 b = -6 c = -b = - ( -12 ) = 12 x = 2c = 2 . 12 = 24 5x = 5 . 24 = 120 4. 2p – 1 = 11 2p p
= 12 =6
1
15−14
10. Ax = B
Uji Kompetensi 4 . 5
c = 2b = 2 . 4 = 8
−1 − 6 −3 − 6 5 )=[ ] −2 + 8 −6 + 8 −3 4 −19 =( )(B) −7 17
−2 3 1 )( 1 −4 2
2q + 3 = -9 2q
= -12
q
= -6
= ( -2 + 12 + 0 ) – ( -2 + 16 + 0 ) = 10 – 14 = -4 ( B )
2r + 1 = 5 2r
=4
r
=2
13. Matrik singular
D=0 -15x – 30 = 0 -30 = 15x
p + q + r = 6 + (-6) + 2 = 2 ( C )
-2 = x
2 −1 12 4 5 8 )−( ) + 2( ) 3 7 −8 9 −6 2 −3 −9 24 8 =( )+( ) 9 5 −16 18 21 −1 =( ) (A) −7 23 −7 −1 −2 6 11 −13 6. ( )+( )−( ) −5 −10 −9 12 −4 3 −20 18 =( ) (B) −10 −1 5. (
2 −4 ] −6 8 1 2 −4 = [ ] −8 −6 8
14. A-1 =
1
16−24
[
1
1
− 4 =[ 3
2
−1
4
x
15. [y] = Kunci jawaban Matematika kelas x
(E)
32
1 6−16
[
]
(D)
6 4 −3 ][ ] 4 1 2
Sagufindo Kinarya
−18 + 8 ] −10 −12 + 2 1 −10 = [ ] −10 −10 =
1
x 1 [y] = [ ] 1
x
a=4
[
6b = a + 8 6b = 4 + 8 6b = 12
(E)
b=2
3 −2 2 3 p ][ ][ ] 1 1 5 −2 q 6 − 10 9 + 4 p =[ ][ ] 2+5 3−2 q −4 13 p =[ ][ ] ( C ) 7 1 q
2c = 3a – b
16. [y] = [
17. A-1
=3.4–2 2c = 10 c=5 −1
4 2 2 1 21. 2 [ 2 ] + 3 [0] + k [1] = [−3] 1 3 3 −2 −2
Bt
=
−2 1 𝑎 −𝑏 [ ]=[ ] −8−3 −3 4 𝑥 𝑐 1 −2 1 𝑎 −𝑏 [ ]=[ ] −5 −3 4 𝑥 𝑐 1
2
[53 5
b=
− −
−2 12 2 2 [ 1 ] + [ 0 ] + k [1] = [−3] +1 9 3 −2 10 2 2 [ 1 ] + k [1] = [−3] +10 3 −2 10 2 2 k [1] = [−3] − [ 1 ] +10 3 −2 2 −8 k [1] = [ −4 ] 3 −18
1 5 4]
=[
5
1
𝑎 𝑥
−𝑏 ] 𝑐
(B)
5
k = -4
18. Det A = Det B
−1 − 6 −3 − 6 1 −3 −1 −3 22. [ ][ ]=[ ] −2 + 8 −6 + 8 2 4 2 2 −7 −9 =[ ] (B) 6 2 1 2 1 2 3 23. [ ] [3 4] = 4 5 6 5 6 1 + 6 + 15 2 + 8 + 18 22 28 [ ]=[ ] 49 64 4 + 15 + 30 8 + 20 + 36
2x2 – 9 = 10x + 3 2x2 – 10x – 9 – 3 = 0 2x2 – 10x – 12 = 0 X2 – 5x – 6 = 0 (x–6)(x+1)=0 x = 6 √ x = -1
(A)
3 4 1 3 4 19. |2 x 5| 2 x 3 2 3 3 2
(A) 1 4 −2 24. 𝐴−1 = 4−6 [ ] −3 1 −1 4 −2 = 2 [ ] (A) −3 1
Det = 0 ( 6x + 60 + 4 ) – ( 3x + 30 + 16 ) = 0 ( 6x + 64 ) – ( 3x + 46 ) = 0
=
25. 3+ 4y = 11
6x + 64 – 3x – 46 = 0
4y = 8
3x
y=2
= -18 x
20. 2A
(E)
= -6 ( E )
5+z+y=4
Bt
5+z+2=4
2a + 2 a + 4 5 a ]=[ ] a + 8 3a − b 3b c 2𝑎 + 2 𝑎 + 4 10 2a [ ]=[ ] 𝑎 + 8 3𝑎 − 𝑏 6b 2c
z
2[
3x – z + 2
= 14
3x – (-3) + 2 = 14 3x
2a = a + 4 Kunci jawaban Matematika kelas x
= -3
33
=9 Sagufindo Kinarya
x
=3
C. S
x + y + z = 3 + 2 + (-3) = 2 ( D )
D. B E. B
Uji Kompetensi 6 . 1 Uji Kompetensi 6 . 2 1. A. Pernyataan Benar
1. A. 1. 2 + 3 = 5
B. Bukan Pernyataan
2. 5 adalah bilangan ganjil
C. Pernyataan Benar
B. 1. Hari ini mendung
D. Pernyataan
2. Hari ini gerah
E. Pernyataan Salah
C. 1. 4 x 5 = 20
2. A. Ada siswa SMK tidak rajin belajar
2. 20 bilangan prima
B. Dua bilangan prima
D. 1. Mas Boi mencintai Iteung
C. Semua murid yang menganggap
2. Iteung tidak mencintai Mas Boi
matematika itu mudah.
E. 1. 3 bukan bentuk akar
D. Ada demonstran tidak memakai ikat
2. 3 = √3
kepala. E. Kuadrat setiap bilangan real tidak selalu
2. A. B ∧ S = S
tak negative.
B. B ∧ B = B
3. A. S
C. S ∧ S = S
B. S
D. B ∧ B = B
C. S
E. B ∧ S = S
D. B
3. A. Saya ingin belajar dan naik kelas
E. B 4. A.
B.
x2
B. Saya ingin belajar tetapi tidak naik kelas + 7x + 10 = 0
C. Saya tidak ingin belajar tetapi naik kelas
(x+5)(x+2)=0
D. Saya tidak ingin belajar atau naik kelas
x = -5 √ x = -2
E. Saya tidak ingin belajar dan naik kelas
2log
x=4
2x
= 11+ 3
= 16
2x
= 14
x= C. 32x-1
=
4. A. 2x – 3 = 11
24
33
x
2x – 1 = 3 2x
=4
x
=2
=7
B. X2 – 4x + 4 = 0 ( x – 2 )( x – 2 ) = 0 x=2√x=2
D. 3x + 7 = 31
C. 2 + x = 10
3x = 24
x=8
x=8
D. 3log x = 2
E. 4x – 12 ≤ 0
x = 32
≤0
=9
4x x
≤3
5. p q p ∧ q ∼p ∼q ∼( p∧ q ) ∼p∧∼q
5. A. B
B B
B
S
S
S
S
B. S
B S
S
S
B
B
S
Kunci jawaban Matematika kelas x
34
Sagufindo Kinarya
S B
S
B
S
B
S
D. p
q
r
∼q
∼q∧r
S S
S
B
B
B
B
B
B
B
S
S
B
B
B
S
S
S
B
6. A. S V B = B
B
S
B
B
B
B
B. B V B = B
B
S
S
B
S
B
C. B V S = B
S
B
B
S
S
S
D. S V S = S
S
B
S
S
S
S
S
S
B
B
B
B
S
S
S
B
S
S
7. A. 2x – 3 = 11 2x
= 14
p∧ (∼q v r )
9. A. ( p ∧ q ) v r
x=7 B. X2 – 4x + 4 = 0
B. ( p v r ) ∧ ( q v s )
( x – 2 )( x – 2 ) = 0
C. ( q v r ) v p
x=2
8. A. p
q
r
∼q
p∧∼q
B
B
B
S
S
B
B
B
S
S
S
S
B
S
B
B
B
B
B
S
S
B
B
B
S
B
B
S
S
B
S
B
S
S
S
S
S
S
B
B
S
B
S
S
S
B
S
S
B. p
q
r
∼q
∼q∧r
B
B
B
S
S
S
B
B
S
S
S
S
B
S
B
B
B
B
B
S
S
B
S
S
S
B
B
S
S
S
S
B
S
S
S
S
S
S
B
B
B
S
B. ~p
q
S
S
S
B
S
S
C. ~p
~q
C. p
q
r ∼p ∼q ∼pv∼q (∼pv∼q )∧ r
B
B
B
S
S
S
B
B
S
S
S
B
S
B
S
B
S
S
S
B
S
(p∧∼q) v r
10. A.
q
P
B.
P
𝑞 r
p
p∧ (∼q v r )
r
C.
q
s
Uji Kompetensi 6 . 3 1. A. p
q
2. A. S
B=B
S
B. S
S=B
S
S
C. B
B=B
B
B
B
D. B
B=B
S
B
B
S
E. S
B=B
B
B
S
B
B
F. S
S=B
B
S
B
S
B
S
S
S
B
B
B
B
B
3. A. p
S
S
S
B
B
B
S
B
Kunci jawaban Matematika kelas x
r
35
~p
p v ~p
S
B
Tautology Sagufindo Kinarya
S
B
B
B B B
S
S
B
B. p
q
pΛq p
B B S
S
S
B
B
B
B
B
B S B
B
B
B
B
S
S
S
B S S
B
B
S
S
B
S
B
S B B
S
S
B
S
S
S
B
S B S
S
S
B
C. p
q
pvq p
S S B
B
S
B
B
B
B
B
S S S
B
S
B
B
S
B
B
S
B
B
B
B B B S S
B
S
S
S
S
S
B
B B S S B
B
B
B
B S B S S
S
S
B
B S S S B
S
S
B
pvq
B. p q r ∼p ∼r ∼pvq q∧∼r ∼pvq
Taulogi
q
B
B
B
B
B
S B B B S
B
S
S
B
S
B
S
S
S B S B B
B
B
B
S
B
B
S
S
S S B B S
B
S
S
S
S
S
B
B
S S S B B
B
S
S
q ~(p
p
q (p v q)
(p
q) ~p ~q ~q
q)
q∧∼r
4. p
5. p q
pvq p
pΛq
~p
9. p q ∼p ∼q p
q ∼pv∼q ~q
~p p
B B
B
S
S
S
B
B B S
S
B
B
B
B
B S
S
B
S
B
S
B S S
B
S
S
S
S
S B
B
S
B
S
B
S B B
S
B
B
B
S
S S
B
S
B
B
B
S S B
B
B
B
B
B
~p
~q
p ∧ ∼q
p
q∧q
B
S
B
B
B
S
S
S
S
B
S
B
6. A. ~q
~p
B. p Λ ~q
p
A. p
q ≡ ∼p v ∼q
B. p
q ≡ ~q
C. p
q ≡ (p
~p q) ∧ (q
p)
7. A. Sx = 25 x=5
10. A. Ahmad tidak merokok atau ia batuk
B. X3 = 8
Jika Ahmad tidak batuk maka ia tidak
X=2
merokok
C. X – 1 = 3x – 9
B. Hujan tidak turun atau halaman basah
-1 + 9 = 3x – x
Jika halaman tidak basah maka hujan
8 = 2x
tidak turun
4 =x Uji Kompetensi 6 . 4 8. A. p q
r ∼q p∧∼q
Kunci jawaban Matematika kelas x
(p∧∼q)
r 36
Sagufindo Kinarya
q
1. A. Konvers = Jika Warman lari pagi maka hari libur
q ~(p
p ∧∼q
q
= jika hari tidak libur maka
B
B
B
S
S
S
Warman tidak lari pagi.
B
S
S
B
B
B
Kontraposisi = Jika Warman tidak lari
S
B
B
S
S
S
pagi maka hari tidak libur .
S
S
B
S
B
S
Invers
p
q) ~ q
3. p
Kesimpulan : ~( p
B. Konvers = Jika Dulkamdi lulus tes maka
q ) ≡ p ∧ ∼q
ia belajar. Invers
= Jika Dulkamdi tidak belajar
4. A. Besi bukan logam atau tidak keras
maka ia tidak lulus tes.
B. Kotak tidak dapat hidup di darat dan di
Kontraposisi = Jika Dulkamdi tidak lulus
darat
tes maka ia tidak belajar.
C. Guru tidak datang dan murid tidak
C. Konvers = Jika Rosi disenangi oleh guru
senang
maka ia berwajah cantik. Invers
D. Tuti tidak cantik atau tidak pemalu
= Jika Rosi tidak berwajah cantik
E. Tuti manis atau tidak cantik
maka ia tidak disenangi oleh
F. Ia tidak di rumah dan tidak masuk
guru.
sekolah
Kontraposisi = Jika Rosi tidak disenangi
G. Hujan turun dan air tidak meluap
oleh guru maka ia tidak
H. Ia tidak naik kelas dan ia diberi hadiah
berwajah cantik.
I. Ia tidak sehat jika dan hanya jika dia tertawa – tertawa
D. Konvers = Jika Rosiman seorang anggota DPR maka ia seorang anggota MPR. Invers
= Jika Rosiman bukan seorang
Uji Kompetensi 6 . 5
anggota MPR maka ia bukan anggota
1. A. p
DPR.
~q
Kontraposisi = Jika Rosiman bukan
∴~p
seorang anggota DPR maka ia
Ia tidak sakit
bukan seorang anggota MPR. 2. A. Konvers : ~q
∴~p ~p
Amir tidak belajar
p
C. p
Invers : ~(p Λ q)
~r
Kontraposisi : ~r
~(p Λ q)
C. Konvers : ~(~q Λ r) Invers : ~p
Invers : ~( p
p
uang
(p
q)
q)
r
Kunci jawaban Matematika kelas x
r Jika Ali tamat SMK mak ia dapat
~p
~(p
r
∴p
(~q Λ r)
Kontraposisi : r
q
q
Kontraposisi : (~q Λ r) D. Konvers : ~r
q
~q
q
Kontraposisi : q B. Konvers : r
B. p
p
Invers : ~p
q
2. A. Valid B. p
q
~p
tidak Valid
∴~q
q) 37
Sagufindo Kinarya
10. Ada orang tidak berdiri ketika tamu agung C. p
q
memasuki ruangan.
q
tidak Valid
11. Jika sungai tidak banyak ikan maka sungai
∴P 3. p
tidak dalam ( E )
q
p
12. ~( p ∧ ∼p )
r
∼p
( ∼p v q )
q
~r
~r
∴ ~𝑝
∼p
(C)
13. Jika ia seorang eropa maka ia seorang
Ia tidak rajin belajar
Belanda .
(E)
14. p ∧ ~(p Vr)
4. p
q
q
r
p
r
r
s
r
s
p
p∧~q∧~r p
p
s
~P
~q ∧ ~ r ( D )
p
15. Jika x ≤ 2 maka 2x + 1 ≤ 5
∴s
16. Air laut tenang dna nelayan tidak melaut
Kelelawar berkaki dua
mencari ikan
5. A.
q ≡ ∼q
17. p
B. p v q q
~p
r
∴
(D)
p
r
C. p v q
r
p
r
∴p
r
Tidak Valid
19. p ∧ ∼q Perang terjadi dan ada orang tidak gelisah .
q
(E)
p
∴q
(B)
18. A
q
~p
∼p
20. –
Tidak Valid
∴q
21. B
D. –
22. C 23. E 24. p
q
Uji Kompetensi Materi Pokok 06
q
r
p
r
1. D
r
s
r
s
2.
q
∼p
B
B
S
S
B
S
B
B
S
S
B
S
S
B
S
B
B
S
B
S
B
S
S
B
B
B
B
B
p
∼q p q p
q ∧∼q {(p q) ∧∼q}
∼p
∴p
Jika Dinda rajin belajar maka ia bahagia . (B) 25. p
5.
∴∼ p ( B )
[ p v ( q Λ r ) ] Λ [ (~p Λ ~q ) v ~r ] p
q
p
q
∼q
3. C atau E 4.
s
q ∼q (p q) ∧∼q
{(p
q) ∧∼q}
B B
B
S
S
B
B S
S
B
S
B
S B B S
S
B
S S B B
B
B
(C) ∼p
26. p ∧ ∼q Nafila tidak rajin belajar dan ada temannya senang. ( D )
6. B 7. x2 = 25 dan x ≠ 5 (D) 8. E
27. p
q
∼q
q ∼q ∧ p ( p
B
B
S
B
S
B
S
B
S
B
p
q)v
9. E Kunci jawaban Matematika kelas x
38
Sagufindo Kinarya
S
B
S
B
S
S
S
B
B
S
∴p
r (A)
46. Ada peserta ujian nasional yang tidak lulus
( ∼q ∧ p )
(B)
B
47. p
q
B
∼q
B
∴∼ p ( B )
B
48. Jika hasil karya manusia tidak baik maka
28. p ∧ q , q B
S S
r , dan r
s
B
S
B
sumber daya manusia tidak baik. ( D ) 49. ∼q
( D ) atau ( E ) q ≡ ∼p v q
29. p
(D)
31. p
50. p
q
q
r
∴p
30. Saya lulus ujian tetapi saya tidak kuliah
∼p ( A )
r (A)
q
∼q ∴∼ p ( B ) 32. E 33. D 34. p
q
∼q ∴ ∼p
(C)
35. Jika ia diduga tidak bersalah maka ia bukan tersangka ( B ) 36. E 37. 3x + 2 = -1 3x = -3 x = -1
(A)
38. – 39. p
q
∼q ∴ ∼p 40.. p r p
(D) q
p
q
∼q
q
∼r
p
p
∼r
p ∴ ∼p
(A)
41. E 42. D 43. ∼(p ∧ q )=∼ p ∧∼ q (B) 44. B 45. p
q
q
r
Kunci jawaban Matematika kelas x
39
Sagufindo Kinarya
Uji Kompetensi 1.1
1
, ,
2 8 16 5
11
8
8
12
16 16 16
B. , 1,
A. 1. 54 x 20 – 640 : ( -8 ) = 1080 + 80
2
,
1
,
11 5
1, 12 , 8
= 1160
4
6
C. ,
2. 750 – 75 x ( -4 ) + 185 : 8
71
,
80
5 10 100
60
,
71
,
100 100 100
4 71 6 , , 5 100 10
= 750 + 300 + 23,125 = 1073,125
4 7 11
16 21 22
3 4
12 12 12
D. , ,
3. ( 300 + 1100 ) : ( -7 ) x 18 – 60
,
6
11 7 4
– 60
= 1400 : ( -7 ) x 18 = -200 x 18
– 60
= -3600
– 60
,
, ,
6
1
2. A. 5
= - 3660 4. 1.200 : ( -40 ) + 35 x 16 – 250
1
66
5
5
13 =
C.
1 =
D.
7 =
5. 8.000 : ( -50 ) – 2.800 + 60 x 9 = - 160 – 2.800 + 540
3
B.
= -30 + 560 – 250 = 280
16
=
3
4 3
7
15
8
8
3
52
7
7
= - 2420 3. A. B. 1. ( 48 – 2 ) x 21 x Rp 4.500,00 = 46 x 21 x Rp 4.500,00
B.
= Rp. 4.347.000,00 C.
2. 8 x 275 – 350 – 250 – 2 x 350 + 98 = 2.200 – 350 – 250 – 700 + 98
D.
= 998
3 2𝑎 16 3
+
−
14 10
+
8
=
2𝑎
11
3
3
3
−12
=
14
14
−10𝑎
𝑎
=
−6 7
−𝑎
=
10
4
=
2𝑎
5
− =
−9 9𝑎
5
10
3. 453.000 : 56 x 5 4. A.
= 8090 x 5
1 3
= 40.450
1
5
5
15
+ = 4
1
3
5
B. 1 + 3
4. ( 20 x 15 + 15 x 20 + 13 x 25 ) : 37 = ( 300 + 300 + 325 ) : 37 =
925
: 37 C.
5. Rp 100.000,00 x 5 – 75 x Rp 6.500,00 = Rp 500.000,00 – Rp 487.500
D.
5 6
1 1
1. A. , ,
1
2 8 16
8
,
2
,
1
8
24
Kunci jawaban Matematika kelas x
1
5
−
9 24
16
4
5
3
5
2𝑘
−
7 4𝑘
+
=5
1 6𝑘
=
48 15
11
=
7
5
15
+
15
1
15
35
8
3
83
15
=
=5
1 +3 = +
=
16 16 16
3
15
20
= E.
16
83
3
− =
= Rp 12.500,00
Uji Kompetensi 1.2
7
8
=
15
= +
=
= 25
3
+
24
=
35 15
+
48 15
8 15 30
12𝑘
−
21 12𝑘
+
2 12𝑘
11 12𝑘 Sagufindo Kinarya
F.
−6
+
6𝑎
5 −8𝑎
B. C. D.
E. F.
1 2 1 5 𝑥
1
6. A. B.
C.
D.
C.
51
5
81
𝑦
𝑧
𝑧
𝑥
31
42
x
1
:
3
21
105
5 10
3
11
4
2
4
4
2
3 3
3
3
1 : = ∶ 5
1
37
8
4
8
F.
7. A.
4𝑏 2 :
2𝑏
3𝑎2 2𝑎 5𝑏3
:
3𝑎2
=
7𝑏
5𝑏3
9
B.
=
4
34
x
82
2𝑏
2𝑎
7
=
11
C.
14
D.
=2 x
41
=
3
4
3
4
= =
4
11 10 4 77 6
(
3
4
+ x
1
1
6
3
3
1
7
1
4
9
4
(3 − 1) = × 2 7
9
7
3
9
4
4
4
11
3
42
+ )=
24
98 9
+
4
−1 19 62
+2
24
93
×
4 5
×
246 5
13
=
2
6
×
+
164
6
1539
5
5
12 10
+
=
5
4
=2( =2x
x
17
×
36 10
10
4
4
22
8
=2(
11
3
= 12
5
147
=
3
5
=9
24 9
884 150
+
21 24
=9
57 4
:
5
208 15
3
48 10
3 5
18 5
)
54
+
)
15
104
24 5
=4
4 5
1
3
3
5
=3 ×3 = =
102 3
×
18 51
36 3
= 12 cm2
15 13
= 13
15
L=pxl
1
3
4
4
=2(5 +2 ) 4
= 2 (7 4) 2
=
L=pxl
3
15
4
342
1
50
150
4
×
=
+
134
=5
153 34
B. k = 2 ( p x l )
6
Kunci jawaban Matematika kelas x
3
98
=2(3 +3 )
10𝑏2
3
3
8
9. A. k = 2 ( p + l )
21𝑎
4
11
9
=
= 18𝑏
3
x
7
17
34
1
10
1
1 (3 − 3 )
= +
3
=1 9
21
37
=
7𝑏
31
x
31
17
37
42
42
21
x
(3 x 2 ) + ( x 2 ) =
8
= 23
x5
513 905 5 = 392 + = = 25 36 36 36 36
11
=
2
∶
=2
9
= 4𝑏 2 x
9
185
7
=6
7
∶
1 2
25
=
1
1
4 :4 =
8. A.
102
3
2 :3 =
8
= × = =1
3
x
51
D.
6
3
x
3
=0
1
1
=
=1
51
2
3
37
6
81
x
x x3
5
8
3
(4 ) =
5
=
93
1
=1 x0
=1
51
x
51
37
=
= E.
24𝑎
=
3
2
B. 4 8 (1 3 + 3 3)
4
4
=
5
26
1
1
1
24𝑎
−
1𝑏 = 𝑏
x
2
24𝑎
−15
6
x1 x
𝑦
12𝑎
+
24𝑎
x1 =
8
−24
=
1
=
3
x
4
3
1
x
13
−65
= 5. A.
−
1
3
4
4
=5 ×2 =
21 4
×
11 4
Sagufindo Kinarya
7 8
231
=2x8
=
= 16 cm
= 14 16 cm2
76.400
x=
16
28.650
× 3 𝑑𝑜𝑙𝑎𝑟 = 7,9 𝑑𝑜𝑙𝑎𝑟
7
3
6. Timah putih 1
1
3
5
10. A. Matematika = 1 − − 15
=
15
7
5 15
−
= 37,5 kg 3
7. A. 1 cm
15
15
1 x=1:
15
=1×
× 21 orang
15 71
2100 𝑚 30 𝑚
8. A. 1 cm 1,5
30 m
cm
x=
: 4
x 4 3
1
× 30 𝑚 = 40𝑚
9. 4 cm
2. A. L1 : L2 = 6S12 : 6S22
6m
1,6 cm
= 6 . 32 : 6 . 42 =9
× 30 𝑚 = 45 𝑚
1
6
22
:
2100 m
B. 1 cm
= S12 : S22 12
30 m
1,5 cm
4
x=
: 16
B. V1 : V2 = S13 : S23 = 33 : 43
1,60,4 41
x
× 6 𝑚 = 2,4 𝑚
10. Ps = Pp : S
= 27 : 64 = 12 cm : 3. 300 ekor 240 ekor x=
300 240
× 1 𝑐𝑚 = 70 𝑐𝑚
× 213 orang
Uji Kompetensi 1. 3
=1
30 m 2100 m
x=
x
=
× 1 𝑐𝑚 = 20 𝑐𝑚
30 𝑚
x
= 45 orang
1. L1 : L2
6002 𝑚
B. 1 cm
x 7
600 m
x=
21 orang
12
30 m
x
7
= B.
−
= 4 × 50 kg
1 400
24 hari
400
= 12 cm ×
x
× 24 ℎ𝑎𝑟𝑖 = 30 ℎ𝑎𝑟𝑖
1
= 4800 cm = 48 m
4. 100 km
2 ,8 ℓ
x x=
2,814 0,2
5. 3 dolar x
ℓs = ℓp : S
0,2 ℓ
= 9 cm :
× 100 𝑘𝑚 = 1400 𝑘𝑚 = 9 cm ×
1 400 400
Rp 28.650,00
= 3600 cm
Rp 76.400,00
= 36 m
Kunci jawaban Matematika kelas x
3
1
Sagufindo Kinarya
11. A. B.
C. D.
3 51
Harga satuan
7 2
75
= 20.000 Laba
× 100% = 2%
100
× 100% =
68
=
12. A. 35 % =
35 100
12,5
B. 12,5 % =
=
C. 3%
B.
13 2 9
8
15
15.
9
115 100
=
23
=1
20
× 260 𝑘𝑚
Uji Kompetensi 1. 4
20
1. A. 34+7 = 311 B. (x + 1)8+6 = (x + 1)14
3
8
5
5
C. 2 (x3×4) = 2x12 𝑎
= 1,6
𝑎
D. ( 𝑏 )5+2 = ( 𝑏 )7 E. ( y3+2 )4 = ( y5 )4 = y5x4 = y20
= 0,73 75
242
90
804
3
= 1,54
25 6 × 100 % =
819
201
= 65 km
F. 14.
x
100
C. 1 = D.
x =
1
= 0,22
11
260 km
20 %
3
D. 115 % = 20
100
× 20.000 = 4000
3
20
=
100
% = 22 %
3
7
=
20
Harga jual = Rp 24.000,00
2
20. 80 %
13. A.
= Rp 2.800.000,00 : 140
× 1004 % = 28%
251
17
19. 7 kodi = 7 x 20 = 140 biji
× 10020 % = 60%
2
𝑎3𝑥4 𝑏 2𝑥4
1
𝑎4 𝑏 3
=
𝑎12 𝑏 8 𝑎4 𝑏 3
= a12-4 b8-3 = a8 b5
% = 37 % 2
2. A. B B. S
× 100% = 90%
C. S D. B
16. Potongan = Rp 11.500,00 – Rp 10.925,00
E. B
= Rp 575,00 % potongan =
575 11.500
F. B
× 100% = 5%
3. A. 6 x 103+5 = 6 x 108 B. 56 x 102 = 5,6 x 103
17. % untung = 100 % - 40 % - 15 % - 7,5 %
C. 960.000 = 9,6 x 105
= 37,5 % Untung =
37,5 100
D. 200 = 2 x 102
× 𝑅𝑝 2.750.000,00
E. 5 x 105-2 = 5 x 103 F. 18.600 = 1,86 x 104
= Rp 1.031.250,00 18. Bonus
1
4. A. ( 24 )4 = 2
15
= 100 × 𝑅𝑝 375.000 = 56.250
2
Gaji pokok
= 375.000
B. ( 33 )3 = 32 = 9
Gaji total
= 431.250
C. ( 82 )2 = 83 = 512
Kunci jawaban Matematika kelas x
3
4
Sagufindo Kinarya
1
1
3
4
F. X3 = 125
D. 4 x ( 43 ) - x ( 34 )
3
X = √125
= 4 x ( 4-1 ) x 3 =4x
=5
1
x3
4
=3 E. (
23
x
2
106
) = (( 2 x 3
102 )3)
6. A.
2 3
= ( 2 x 102 )2 = 4 x 104 1
F. 34 x ( 33 ) : 3 2
=3
B.
1 − 2
C.
3 1 4+2+2
E.
5. A. ( 25 )x = 2-4
1 2
1
𝑞
−3
𝑥
−2
3
𝑧
1 −1 2
=𝑧
−
1 2
7. A. √𝑎
5 x = -4
3
4
B. √𝑏
5
C. √𝑐 2
3
1
5
D. √𝑑 4
B. ( 3-1 )x -2 = 3-2 . 32 3-x+2 = 3
=𝑝
1 𝑝2
−
D. 𝑦 2
= 36
x=−
1
−1
1 2
8.
-x + 2 = −1
1
3 𝑥 103 𝑚 1 𝑑𝑒𝑡𝑖𝑘
=
3000 𝑚
5 = 𝑚⁄𝑗𝑎𝑚 3600 𝑗𝑎𝑚 6
2
9. 2,7 𝑥 10−23 𝑞 𝑥 4 𝑥 1028
1
32 = x
= 10,8 𝑥 10−23+28
C. x = 3 1
= 10,8 𝑥 105 𝑞
D. ( 52 )3x = 52 36x = 5 6x = x=
= 1,08 𝑥 106 𝑞
1 2
1
10. F =
2 1
𝑘 .𝑄.𝑞 𝑟2
𝑟2 =
12
𝑘 .𝑄 .𝑞 𝐹
4
E. ( 32 )x+3 = √(33 )𝑥 + 5 32x+6
=3
2x + 6 =
𝑟=√
3𝑥+15 4
3𝑥+15
= (𝑘
4
.
𝑘 .𝑄 .𝑞 𝐹
𝑄 .𝑞 1 𝐹
)2
8x + 24 = 3x + 15 8x – 3x = 15 – 24 5x = -9 x=−
9 5
Kunci jawaban Matematika kelas x
5
Sagufindo Kinarya
Uji Kompetensi 1. 5
F. 3√7 5. A. 2√15 + 2√10
1. A. Bukan B. Bukan
B. √48 - 2√18 = 4√3 - 6√2
C. Ya
C. (√3 )2 + 2 (√3 ) (√2 + (√2 )2
D. Bukan
= 3 + 2√6 + 4
E. Bukan
= 7 + 2√6
F. Ya
D. √6 x 6√6 = 6 x 6 = 36 E. 5.2 - √6 + 5√6 - 3
2. A. √27 = 3√3
= 10 + 4√6 - 3
B. √28 = 2√7
= 7 + 4√6
C. √25 .3 = 5√3
F. (√5 )2 - 2(√5 ) (√3 ) + (√3 )2
D. √49 .3 = 7√3
= 5 – 2√15 + 3
E. 2√49 .2 = 2.7√2 = 14√2
= 8 - 2√15
F. √25 .2 = 5√2 G. √100 .2 = 10√2
× √7 2 = √7 7 √7 × √7 √3 × √2 √6 B. = 2 √2 × √2 1 × 1−√3 1−√3 1−√3 C. = = 1+√3 × 1−√3 1−3 −2
6. A.
H. √100 .3 = 10√3 I. 5√4 .2 = 5.2√2 = 10√2 J. 10√100.10 = 10.10√10 = 100√10 3. A. c2 = a2 + b2 =
22
+
32
D.
2
2+√2 2−√2
=4 + 9
=
= 13 c = √13 (betuk akar)
× 2+√2 22+2 .2√2+(√2) = 4−2 × 2+√2
4+4√2+2 2
=
6+4√2 2
2
= 3 + 2√2
× √5 1 1 = √5 = × 2,236 = 0,45 5 5 √5 × √5 20 × √2 20 B. = √2 = 10 × 1,41 = 14,1 2 √2 × √2 10 × √6 10 5 C. = √6 = √2 . √3 6 3 √6 × √6
B. bentuk akar
7. A.
C. bukan D. bukan E. bentuk akar D. bukan
1
5
= 3 × 1,41 × 1,732 4. A. √9. 2 + 8 = 310√2 + 8
= 4,07
× √3 6 = √3 = 2 × 1,732 = 3,46 3 × √3 3 3 × √5 5 2,236 E. = √5 = = 1,12 2√5 × √5 10 2 4 × √2 4 4 F. = √2 = × 1,41 = 0,56 5√2 × √2 10 10
B. 5√3 - 2 √16.3 = 5√3 - 2.4√3
D.
= 5√3 - 8√3 = -3√3 C. 2√5 D. 15√13 + 9√13 - 2√3 + 3√3 = 24√3 - √3 2
6
3
E. √ 4 . 6 + √ 64 . 3 + 2√−81 3
= 2√6 + 4√ 3 + 3.9 3
= 2√6 + 4√ 3 + 27 Kunci jawaban Matematika kelas x
6
Sagufindo Kinarya
1
8. A. sekawan = √3 -1
B. x = 3-3 = 27
(√3 + 1) (√3 - 1) = (√3 )2 – 12 = 3 – 2 = 2
C. x4 = 16
B. sekawan = √5 +2
x=2 D. x-3 = 0,125
(√5 - 2) (√5 + 2) = (√5 )2 – 22 = 5 – 4 = 1 C. Sekawan = √7 + 3 (√7 - 3) (√7 + 3) = (√7 )2 – 32 = 7– 9 = - 2
1 𝑥3 1
D. Sekawan = 3 +√2 (3 -√2 ) (3 +√2 ) = 9.
125
x-3 =
𝑥
32
– (√2
)2
=9–2=7
1000 125
=
=
1000 5
10
x=2
L =pxℓ
5
3. A. log 2 × 4 = log 10 = 1
= (2+√2 ) (2-√2 ) B.
= 22 – (√2 )2
5log
155 ×25 62
=4–2
= 5log 125
= 2 cm2
=3
d2 = p2 + ℓ2
C. 3log 72 x 7log 33
= (2 + √2)2(2-√2 )2
2 . 3 x 3log 7 x 7log 3
= (4 + 4√2 +2) + (4 - 4√2 +2)
= 6 x 3log 3
= 12
=6 x 1 =6
d = √12 = 2√3 cm
4. A. 25 = x + 6 32 x + 6
10. AC = √22 + 12
26 x
= √4 + 1 B.
= √5
9
1 2
= 2x – 1
3 = 2x – 1
AD = √𝐴𝐶 2 + 𝐶𝐷2
4 = 2x
2
= √(√5) + 12
2=x
= √5 + 1
C.
= √6
10log
𝑥 16
103 =
5.
1. A. 5 B. 3log
1 − 3 3 .3 2
10log
D. 2log
10-3
3 −− 3 2 .=
= 3log
1 − 33 2
𝑥 16
=
A. log 24 = log (23 x 3 ) = log 23 + log 3
1 − 32
= 3 ( 0, 301 ) + 0, 477
= -3 -
=3
16.000 = x
Uji Kompetensi 1.6
C.
×2
= 0, 903 + 0, 477 = 1, 380
3 2
B. Log 600 = log ( 2 x 3 x 100 ) = log 2 + log 3 + log 100
2. A. x = 42 = 16
= 0, 301 + 0, 477 + 2 Kunci jawaban Matematika kelas x
7
Sagufindo Kinarya
= log 3,981 – log 104
= 2, 778 1
= 0,6 – 4
1
C. Log 2 . 32 = log 2 + log 32 1
= 0, 301 +
2
= -3,4
x 0, 477
= 0, 301 + 0, 2385
9. A. 0,3856
= 0, 5395
B. 3,477 C. -0,6198 1 2
D. -1,28735
D. Log 23 . 32 + log 3 . 5
1
= log 23 + log 32 + log 3 + log 52 10.
1
= 3(0,301)+2(0,477)+ 0,477 + (0,699) 2
6.
=
7.
C. 1479,1
= 2, 6835
D. 0,22
=
=
50 =
𝑙𝑜𝑔2 50
Uji Kompetensi 1.7
𝑙𝑜𝑔2 6
2 log 2 .52 2 log 2 .3
2𝑙𝑜𝑔2+2 log 52
=
1. ( 32 )2x – 1 =
2 log 2+2 log 3
34x-2
1+2 .𝐵
15 =
3 log 6 3 log 15
3 log 2 .3
3 log 2+3 log 3
𝑃+𝑄
35
= 3-5
4x
= -3
x
=
2. (25 )
= 3 log 3+2 log 3
3 log 3 .5 1+𝑃
1
4x – 2 = -5
1+𝐴
6log
=
B. 17,5
= 0, 903 + 1, 431 + 0, 3495
6log
=
A. 2,5
−3
2𝑥−3 2
𝑥+1 4
= (27 )
10𝑥 − 15 7𝑥 + 7 = 2 4
1 +1 𝑝
= ( )× 𝑃 1+𝑄
×𝑃
40– 60 = 14 + 14
1+𝑃
40–14 = 14 + 60
𝑃+𝑃𝑄
26 = 74
8. A. log 39, 81 = log ( 3, 981 x 10 ) = log 3, 981 – log 10
𝑥=
= 0, 6 – 1 = - 0, 4
(B)
4
74 26
=
37 13
(E)
4
B. Log 398100 = log 3, 981 x 105
3. 3
2𝑥−1 3
= log 3, 981 + log 105
2𝑥−1
= 0, 6 + 5
3
= 5,6
= 3−3
= −3
2 -1= -27
C. Log 0, 03981 = log 3, 981 : 102
2=-26
= log 3, 981 – log 102 = 0,6 – 2
= -13
= -1,4 D. Log 3,981 : 104 Kunci jawaban Matematika kelas x
8
Sagufindo Kinarya
2 4𝑥−2
4. ( ) 3
2 0
= 74 + 14√7 (E)
=( ) 3
4 -2 = 0
3
2
2
3
3
11. √54 + √16 − √250
1
= 4 = 4 = 2 (C)
3
3
3
= 3√2 + 2√2 − 5√2 = 0 (E)
5. P = 3 ℓ
d2 = p2 + ℓ2
12. 7log 23.2log32.3log 7-1 )2
L = 72
= ( 3√24
P x ℓ = 72
= 216 + 24
3ℓ x ℓ = 72
= 240
3ℓ2
+ 24
= 3.2(-1).7log2.2log 3.3log 7 = - 6. 7log 7 = - 6.1
d = √240
= 72
ℓ2 = 24
= -6
= 4√15
(C)
ℓ = √24 13.
p = 3ℓ
log 98 log 6
= 3√24 = 6.
r4
r3
x
=
r4+3-1
:
r1
= r6 = 2
4
=
log2 2 . 3
1+2 .2 log 7 1+𝑏
1+
=
log2 2 + log2 72 log2 2 + log2 3
2 log7 2
1+𝑏
2 𝑎
×𝑎 1+𝑏 ×𝑎 1+
=
log 2 .72
=2
𝑎+2
(C)
𝑎(𝑏+1)
1
7. (5−3 )−3 + (23 )3 − (103 )3 = 52 + 24 -10
14.
log5 75
=
log5 5
= 25 + 16 + 10
=p+2 4
=
9.
×1 − √5 4 ( 1− √5 ) = 12 −5 − √5
1+√5 ×1
4 ( 1−√5) −4
= −1 + √5
15. 3log 3
(A)
125 5
log 25
3
2−√3×2+√3
(D) 10
: 3log
2
log 25 : 3log 5
3
2 ×2+√3
=
1
= 5log 3 + 2 . 5log 5
= 31 (D)
8.
log5 3 . 52
= 3log 25 = 2 (A)
log 5
2(2 + √3) 2
22 − (√3)
16. 3log
41 . 54 27 82
= 3log 27 = 3 ( C )
= 4 + 2√3 (B) 17. 2log 5 x 5log 6 – 2log 24 = 2log 6 – 2log 24
10. (3√7 + 5)(4√7 − 2)
= 2log
= 12.7 − 6√7 + 20√7 − 10
2
24 1
= log
= 84 + 14√7 − 10 Kunci jawaban Matematika kelas x
6
4
9
Sagufindo Kinarya
= 14.000.000 cm
= -2 (A)
= 140 km
(B)
3
log 81
18. 15log 81 = = = =
4 3
log 3 × 5
3
log 15
=
9 3 35 25. 21 . - 2. + 42 2 8
4 3
log 3+ 3log 5
=
4
9 2
3
1+ log 5 4 1 1+ 𝑃
=4
×𝑃 4𝑃 = ×𝑃 𝑃+1
(B)
-
9
3
+4
2
8
3 8
26. ( E )
19. H Jual = 115.000 x 100 = 11.500.000 H Beli =
= 10.500.000 _
27. ( B )
Untung
= 1.000.000
28. L = 𝜋 r 2
1.000.000
% untung =
10.500.000
= 9,5 %
20.
V1
: V2
S13
: S23
23
r2 =
x 100 %
(E)
𝜋
r =√
. 24 . 24
1: 8
(D)
6010 366
= 3 √2 + 4√2 - 4√2 + 5√2 - 6 √2 = 2 √2
x 12
2
orang
=
2 10
= 24
= - 13
x 120 dm
(D)
33. P = 8 . ( 4-2 )1/2 . ( 64 )-1/3 = 8 . 4-1 . ( 43 )-1/3
x 120 dm
= 82 .
(C)
=
24. ∝s = ∝p : s = 7 cm :
( B )
32. 3 + 4√3 - 4 √3 - 16
= 20 orang ( E ) 2+3+5
( B )
31. 3 √2 + 2√4 . 2 - √16 . 2 + √25 . 2 – √36 . 2
(B)
36 hari x
2
( D)
𝑔
x 800 gr
250
22. 60 hari 12 orang
23.
𝐿
= 7 √15
= 1200 gr
x=
𝐿 𝑔
30. √735 = √15 .49
375 gr x 375
(D)
𝜋
= 2𝜋 √
21. 250 gr 800 gr
x=
𝐿
29. T = √4𝜋 2 .
: 43
12. 12. 12 : 2 4
𝐿
1
41
.
1 4
(D)
2
34. ( α2 + b2 )- ½ =
2.000.000
= 7 cm x 2.000.000 Kunci jawaban Matematika kelas x
1
1
= 10
1 √α2 + 𝑏2
1 (𝛼 2 + 𝑏2 )1/2
( C) Sagufindo Kinarya
35. 1 mol 1 2
6,02 . 1
mol
=
1023
x 60,2 . 1022
20
= 3,01 x 1022 7
24
36. (
)
3 .5
=
228 37 . 57
:(
312
x
212 . 53
x
=22 . 1013
23 . 3 52
9
5
. 3o
= 2.
(D)
3
44. 2. 4 − 3. 2 +
)
(D)
35 8
18 9 35 − + 4 2 8
510
36−36+35
215 . 35
8
= 228 – 12 15 . 312 – 7 – 5 . 510 – 3 – 7 = 21
. 26 . 108
27
= 23+6-7 . 105+8 ( E )
3
4
3 x( 4 ) 2 .5
105 . 23
=
35 8
3
= 4 8 (D)
45. E
. 5o Uji kompetensi 2.1
37. alog b . blog c . clog d = alog d
(A)
1. membilang = sesuatu yang eksak banyak siswa dalam suatu kelas
×3 − √5 4 ( 3− √5 ) 38. = 3+√5 ×3 − √5 9−5 4
)2
= ( 3 - √5
=
– 2 . 3√5 +
32
mengukur = pendekatan panjang , luas (√5)2
= 9 – 6√5 + 5
2. tiga cara :
= 14 – 6√5
1. pembulatan ke satuan terdekat
(D)
2. pembulatan ke angka desimal 39. log 23 . 3 . 53 = 2 log 2 + log 3 + 3 log 5 = 2 A + B + 3C
40. log ( x – 5 ) = log x-5 =
3. pembulatan ke banyaknya angka yang signifikan (berarti)
(D)
𝑥
3. A. 10, 84
5
B. 5, 42
𝑥
C. 157, 22
5
D. 1072 , 12
5x – 25 = x 4x =
25 4
=6
41. V = a
4. A. 0,1246 B. 0,008543
1
C. 172, 5
(B)
4
D. 1,281 5. A. 3 angka signifikan
d = s√3 3
S3 = a
B. 4 angka signifikan
= √𝑎 . √3 6
3
= √27𝑎2
S = √𝑎
C. 6 angka signifikan
(C)
D. 5 angka signifikan 42.
100.000 2.500.000
x 100% = 4 %
(C)
6. A. 17 B. 171
43.
105 27
10 3
26 . 104
( ) .( 5
23
C. 1708
2
)
Kunci jawaban Matematika kelas x
D. 17085 11
Sagufindo Kinarya
14. A. SM = ½ x 0,01 = 0,005 ≈ 0,01 kg 7. A. 45,1 SR =
B. 503
0,05
5
=
7,25
725
C. 0,1 % kesalahan =
D. 0,6
1
=
145
1 145
x 100% = 0,69
B. SM = ½ x 1 = 0,50 kg 8. A. membilang jumlah murid kelas X-1
SR =
B. mengukur panjang meja memakai penggaris
0,50 98
5
=
=
980
% kesalahan =
B. 10
SR =
C. 5,124 D. 199
0,5
1 196
100
5
=
1
1 200
200 x 100% = 0,50 %
D. SM = ½ x 0,1 = 0,05 menit
B. 1,71
SR =
0,05 9,5
5
=
950
=
C. 1,71 % kesalahan =
D. 1,714
11. Salah mutlak = ½ x satuan ukur terkecil Salah relatif =
𝑠𝑎𝑙𝑎ℎ 𝑚𝑢𝑡𝑙𝑎𝑘
SR =
ℎ𝑎𝑠𝑖𝑙 𝑝𝑒𝑛𝑔𝑢𝑘𝑢𝑟𝑎𝑛
1 km
749,5
B. 53,2 jam 0,1 jam 0,5 53,25
53,15
C. 8,01 detik 0,01
0,005 8,015
7,995
D. 45 cm
0,5
44,5
45,5
SR =
Salah relatif =
=
45,08
5 45080
=
8919
=
SR = 1
9016
25,10
=
89190
2510
=
1
37,05
=
x 100% = 0,53 %
1
=
154
1 154
x 100% = 0,65 %
0,05 3
=
5 30
1
= 1 6
6
x 100% = 16,67 %
0,005 90,05
=
5 90050
=
1 18010
1 18010
x 100% = 0,006 %
D. SM = 0,1 x ½ = 0,05 menit
17838
SR =
=
1 502
5 37050
Kunci jawaban Matematika kelas x
=
0,05 8,6
=
5 860
% kesalahan =
D. SM = 0,01 x ½ = 0,005 0,005
190
≈ 0,01 %
5
5
770
% kesalahan =
C. SM = ½ x 0,1 = 0,05 0,05
1
C. SM = 0,01 x ½ = 0,005 m
B. SM = ½ x 1 = 0,5 0,5
77
5
=
% kesalahan =
13. A. salah mutlak = ½ x 0,01 = 0,005 0,005
190
B. SM = ½ x 1 = 0,5 liter
0,5 750,5
1 cm
0,5
% kesalahan =
12. Pengukuran terkecil SM atas bawah A. 750 km
1
15. A. SM = ½ x 1 = 0,5 ton
persentase kesalahan = salah relatif x 100%
SR =
x 100% = 0,51
=
1000
% kesalahan =
10. A. 1,7
SR =
196
C. SM = ½ x 1 = 0,50 gr
9. A. 0,0018
SR =
1
=
1 172
1
172
x 100% = 0,58 %
1 7410
12
Sagufindo Kinarya
Uji kompetensi 2.2
SM = 0,005
C. pengukuran 1 = 5,51 gr
pengukuran 2 = 10,41 gr SM = 0,005 ------------------------------- + ----------------- + = 15,92 gr SM = 0,01
1. A. Pengukuran terbesar = 17,5 gr Pengukuran terkecil = 16,5 gr
Batas-batas pengukuran = 15,92 ± 0,01
B. pengukuran terbesar = 89,5 cm
Pengukuran terbesar =15,92+0,01=15,93
pengukuran terkecil = 88,5 cm
Pengukuran terkecil = `5,92–0,01=15,91
C. Pengukuran terbesar = 90,65 detik Pengukuran terkecil = 90,55 detik
7. A. pengukuran 1 = 15 kg
SM = 0,5 kg
pengukuran 2 = 10 kg SM = 0,5 kg ------------------------------- - --------------- + 5 kg SM = 1 kg
D. Pengukuran terbesar = 48,485 gr Pengukuran terkecil = 48,475 gr
Batas-batas pengukuran = ( 5 ± 1) kg
2. A. 8,8 – 8,3 = 0,5 cm
Pengkuran terbesar = 5 + 1 = 6 kg
B. 9,90 – 9,87 = 0,03 gr
Pengukuran terkecil = 5 – 1 = 4 kg
C. 77,8 – 77,4 = 0,4 kg
B. pengukuran 1 = 8,5 cm SM = 0,05 cm
D. 7 – 5 = 2 km
pengukuran 2 = 5,4 cm SM = 0,05 cm ------------------------------- + ------------------ + 3,1 cm SM = 0,1 cm
3. A. ( 12,5 ± 4,5 ) m B. ( 45,5 ± 0,05 ) kg
Batas-batas pengukuran = (3,1 ± 0,1) cm
C. ( 8,676 ± 0,06 ) mm
Pengukuran terbesar = 3,1 + 0,1= 3,2 cm
D. ( 0,6 ± 0,05 ) cm
Pengukuran terkecil = 3,1 – 0,1 = 3 cm
4. SM = ½ x 1 cm = 0,5 cm
C. pengukuran 1 =20,50 gr SM =0,005 gr
batas atas
= 167 + 0,5 = 167,5 cm
batas bawah
= 167 – 0,5 = 166,5 cm
pengukuran 2 =17,59 gr SM =0,005 gr ------------------------------- + ----------------- + = 2,91 gr SM = 0,01 gr
5. A. pengukuran terbesar = 15,4 + 0,1 = 15,5
Batas-batas pegukuran = (2,91±0,01) gr
pengukuran terkecil = 15,4 - 0,1 = 15,3 8. Keliling maksimum yang mungkin
toleransi = 15,5 – 15,3
K = 2 ( 12,5 + 15,5 )
= 0,2
= 56 cm
B. Toleransi = 75,17 – 75,13 = 0,04
Keliling minimum yang mungkin
C. Toleransi = 103,9 – 103,7 = 0,2
K = 2 ( 11,5 + 14,5 ) 6. A. pengukuran 1 = 8 kg
SM = 0,5 kg
= 52 cm
SM = 0,5 kg + + 23 kg SM = 1 kg Batas pengukuran = ( 23 ± 1) kg pengukuran 2 = 15 kg
9. Luas maksimum yang mungkin L = 10,5 x 7,5 = 78,75 cm2
Pengkuran terbesar = 23 + 1 = 24 kg
Luas minimum yang mungkin
Pengukuran terkecil = 23 – 1 = 22 kg
L = 9,5 x 6,5
= 61,75 cm2
B. pengukuran 1 = 15,1 cm SM = 0,05 10. Keliling maksimum
pengukuran 2 = 22,2 cm SM = 0,05 + + 37,3 cm SM = 0,1
K = 4 x 5,1
Keliling minimum
Pengkuran terbesar = 27,3+0,1= 27,4 cm
K = 4 x 4,9
Pengukuran terkecil = 27,3 - 0,1=27,2 cm Kunci jawaban Matematika kelas x
= 20,4 m
= 19,6 m
Luas maksimum 13
Sagufindo Kinarya
= 26,01 m2
L = 5,1 x 5,1
15. √30,25 = 5,5
Luas minimum
√20,25 = 4,5
L = 4,9 x 4,9
= 24,01
m2
5,5 + 4,5 = 5(D) 2
Uji Kompetensi 2.3
16. pengukuran 1 = 10 cm
SM = 0,5 cm
pengukuran 2 = 6 cm
SM = 0,5 cm
------------------------------ 1. C
------------------- +
= 4 cm
2. D
SM = 1 cm
P. max sisa = 4 cm + 1cm = 5 cm
(E)
3. B 4. A
17. D atau E
5. SR =
6. % =
0,005 3,75
0,05 28,2
=
5 3750
=
1 750
(D)
19. SM = ½ x 0,1 = 0,05
(A)
20. pengukuran 1 = 5,5 m SM = 0,05 pengukuran 2 = 3,8 m SM = 0,05 ------------------------------ ------------------- + = 1,7 m SM = 0,1 m
x 100%
= 0,177%
18. T = 7,6 – 6,8 = 0,8 (C)
(C)
Selisih minimum = 1,7 – 0,1 = 1,6 m
(E)
7. D 8. Toleransi = 375,72 – 375,48 = 0,24 ( C )
21. 24
0,5
7
0,5
9. A
25 +
0,5 +
10. 8, 95
56
1,5
8,87 0,08
56 ± 1,5 ( A ) (C) 1
11. Salah Mutlak = 2 × 0,01 = 0,005 ( D ) 12. pengukuran 1 = 5,7 m
SM = 0,05 m
pengukuran 2 = 3,1 m SM = 0,05 m ------------------------------ ------------------- + = 2,6 m SM = 0,1 m
0,5
8
0,5
8
0,5
8
0,5
32
2,0
32 ± 2,0 ( C ) 23. 3,2
Batas pengukuran terbesar = 2,6 + 0,1 = 2,7 m
22. 8
(C)
0,5
5,4 +
0,5 +
8,6
0,1
4,3 13. 125
0,5
245
0,5
143 +
0,5 +
513
1,5
4,3 ± 0,1 ( D ) 24. L max = 17,5 x 13,5 = 236,25 Lmin = 16,5 x 12,5 = 206,25
(D)
25. 25,5 x 15,5 = 395,25 ( D )
513 – 1,5 = 511,5 ( A ) 14. 8,45 x 6,45 = 54, 5025 ( B )
Kunci jawaban Matematika kelas x
14
Sagufindo Kinarya
1
Uji kompetansi 3.1
< 𝑥
2
∴ HP = { 1,2,3,… }
= 5𝑥 – 3
1. A. 4 + 10
=5𝑥 -4𝑥
10 + 3
3. A. 3 𝑥 + 2 𝑥 ≤ 7 𝑥 + 10
13 = 𝑥
3 – 5 + 6 ≤ 7 𝑥 + 10
∴ HP = { 13 } B. 4 + 4 𝑥
– 10 ≤ 7 𝑥 – 5 𝑥
=2𝑥 -6
– 10 ≤ 2 𝑥
4𝑥-2𝑥 = -6–4
– 10 ≤ 𝑥
2 𝑥 = - 10
∴ HP = { - 4 , - 3 , - 2 ,… }
𝑥 =-5 B.
∴ HP = { - 5 } C. 6 𝑥 - 3 + 1= 4 𝑥 – 2
1−2𝑥
−3<0
2−𝑥 1−2𝑥
6𝑥-4𝑥 =-2+3–1
−
2−𝑥
3(2−𝑥)
1−2𝑥−6+3𝑥
2𝑥 =0
2−𝑥
𝑥 =0
−5+𝑥
∴ HP = { 0 }
2−𝑥
D. 5 (2 𝑥 - 1) = 2 (𝑥 + 1) 10 𝑥 - 5
=2𝑥 +2
10 𝑥 - 2
=2+5
<0
<0
( -5 + x ) ( 2 – x ) < 0 √
x=5
+
7
2
8
∴ HP = {
7 8
}
6 𝑥 + 72
= 3 𝑥 – 27
6𝑥 -3𝑥
= – 27 – 72
3𝑥
= – 99
𝑥
= – 33
C.
4. A.
2𝑥−8
2 3
9𝑥+9−4𝑥+16
6𝑥−24
1
∈𝑅}
− >0
3(2𝑥−8)
>0
>0
(5𝑥 + 25)(6𝑥 − 24) > 0
2
x = -5
∴ HP = { 6,7,8,… }
+
B. 3 + 10 𝑥 -5 < 18 𝑥 – 6
8
3𝑥+3
,𝑥
3(3𝑥+3)−2(2𝑥−8)
2 𝑥 > 11
4
< x
Hp = { | 𝑥 > 3
3𝑥–𝑥 >5+6
4
<0
4
3
2. A. 3 𝑥 - 6 > 𝑥 + 5
3–5+6
3−𝑥
3–x<0
= { – 33 }
𝑥 > 5
5
Hp = { | 𝑥 < 2 atau 𝑥 > 5 }
E. 6 (𝑥 + 12) = 3 (𝑥 - 9)
∴ HP
x=2
-
8𝑥 =7 𝑥 =
<0
2−𝑥
-5
< 18 𝑥 – 10 𝑥
x=4 4
Hp = { | 𝑥 < -5 atau 𝑥 > 4 , 𝑥 ∈ 𝑅 }
< 8𝑥 < 𝑥
Kunci jawaban Matematika kelas x
15
Sagufindo Kinarya
B. 3 (7 - 4𝑥) ≤ 𝑥 + 1
HP = {
21 – 12𝑥 ≤ 𝑥 + 1 21 – 1 20
,1 }
( 2x – 1 ) ( x – 1 ) = 0
≤ 13𝑥
x=1 √ x=2
20 ≤ 𝑥 13 7 1 13 ≤ 𝑥
HP = { 1, 2 } G.
7 ∴ HP = { | 𝑥 ≥ 1 13
,𝑥
∈𝑅}
4+𝑥
=
4𝑥
3 𝑥+2
4x + 8 + x2 + 2x = 12x X2 + 2x + 4x – 12x + 8 = 0
C. 5 ( 2 𝑥 – 1 ) ≥ 3 ( 2 - 3 𝑥 )
X2 – 6x
≥ 6–9𝑥
+8=0
(x–4)(x–2)=0
10 𝑥 + 9𝑥 ≥ 6 + 5
x=4√ x=2
19 𝑥 ≥ 11
HP = { 2 , 4 }
11
𝑥 ≥
2
F. X2 – 2x + 1 – x + 1 = 0
≤ 𝑥 + 12 𝑥
10 𝑥 - 5
1
19
∴ HP = { | 𝑥 ≥
11 19
2. A. x2 – 4x
,𝑥
∈𝑅}
=0
X2 – 4x + 4 = 4 ( x – 2 )2
Uji Kompetensi 3.2 1. A. ( 𝑥 - 3 ) ( 𝑥 - 10 ) = 0
=4
x–2
= √4 = ±2
x1
=2+2=4
x2
= -2 + 2 = 0
HP = { 0 , 4 }
𝑥 = 3 √ 𝑥 = 10 ∴ HP = { 3 , 10 }
B. x2 + 3x
B. 3 𝑥 ( 3 – 𝑥 ) = 0
= 10 9
𝑥 =0√ 3=𝑥
(x+
∴ HP = { 0 , 3 } C. ( 𝑥 - 5 ) ( 𝑥 + 5 ) = 0
x+
3 2
)2 =
3 2
∴ HP = { - 5 , 5 }
x1
=
x2
=
D. 0 = 2 𝑥2 - 15𝑥 + 28 0 =(2𝑥 -7)(𝑥 -4) 2𝑥 = 7 √ 𝑥 = 4
49 4
=±
𝑥 =5 √ 𝑥 =-5
𝑥=
9
X2 + 3x + 4 = 10 + 4
7 2
7 2
3
4
2
2
− = =2 7
3
−10
2
2
2
− − =
= −5
HP = { -5 , 2 }
7 2
∴ HP = { 3
1 2
C. x2 – 2x ,4}
=3
X2 – 2x + 1= 3 + 1 ( x – 1 )2 = 4 x–1
=±2
( 2x – 1 )( x – 1 ) = 0
x1
=2+1=3
2x = 1 √ x = 1
x2
= -2 + 1 = -1
E. 2 x2 – 3x + 1 – x + 1 = 0
x =
HP = { -1 , 3 }
1 2
Kunci jawaban Matematika kelas x
16
Sagufindo Kinarya
D. X2 + 1 = 1
3. A. a = 1 , b = 7 , c = 3
X2 = 1 -1
−𝑏±√𝑏 2 −4𝑎𝑐
X1,2 =
X2 = 0
2𝑎
X = ±√0 =
X1 = 0 X2 = 0
= E.
2𝑥 2 +6𝑥=1
X2 + 3x + (x+
9 4
3 2 ) 2
=
x2
=
HP = { −
2
4
7 = 𝑥 2 − 2𝑥 2
4 √11 2
√11 2
3
− = 2
−√11 2
,
√11−3 2
3
−√11−3
2
2
− = √11−3 2
15+√305 8
X2 =
15−√305 8
}
C. a = 2 , b = 1 , c = -6
HP = {
G.
𝑥 2 +2𝑥=3
2 .2
= =
𝑥2 = =
−3√2 2
X2 =
−3√2+2 2
−1±7 4 4
−1−7
6
3
4
2
= =
=
4
−8 4
= −2
D. 7x2 + 13x – 2 = 0 −13±√132 −4 .7 .(−2)
x1,2 =
2 .7
:2 =
2
𝑥 + 2𝑥 + 1 = 3 + 1 ( x + 1 ) = ±2
=
X1 = 2 – 1 = 1 X2 = -2 – 1 = -3
X1 =
HP = { -3, 1 } X2 = Kunci jawaban Matematika kelas x
−1±√49 4
−1±7
X1 =
+1
−3√2+2 3√2+2 , } 2 2
2𝑥 2 +4𝑥=6
−1±√12 −4 .2 .(−6)
x1,2 =
3
= 𝑥1
15±√305 8
X1 =
± √2 = 𝑥 − 1 2
3√2+2 2
2 .4
=
:2
+ 1 = 𝑥1
−(−15)±√(−15)2 −4 .4 .(−5)
X1,2 =
= (𝑥 − 1)2
3√2 2
2
4x2 – 15x – 5 = 0
+ 1 = 𝑥 − 2𝑥 + 1
4
−7−√37
B. 4x2 = 15x += 5
2
18
−7±√37 2 2
X2 =
11
√11−3 2
7=2𝑥 2 −4𝑥
2
9
=±
2
x1
7
1
2 .1
−7±√37
X1 =
= +
=
3
x+
F.
:2
1 𝑥 2 +3𝑥= 2
−7±√72 −4 .1 .3
17
−13±√225 14 −13±15 14
−13±15 14 −13−15 14
= =
2 14
=
−28 14
1 7
= −2 Sagufindo Kinarya
E. a = 1 , b = -2q , c = -q2
= = =
X2 =
−𝑝−√𝑝 𝑝−1
−(−2𝑞)±√(−2𝑞)2 −4 .1 .(−𝑞) 2 .1
4. ( 2x – 2 )2 = x2 + ( x + 2 )2
2𝑞±√4𝑞2 +4𝑞
4x2 – 8x + 4 = x2 + x2 + 4x + 4
2
4x2 – x2 – x2 – 8x – 4x + 4 – 4 = 0
2𝑞±2√𝑞2 −𝑞
2x2 – 12x = 0
2
2x ( x – 6 ) = 0
= 𝑞 ± √𝑞 2 + 𝑞
x=0
X1 = q + √𝑞 2 + 𝑞 √𝑞 2
X2 = q -
√ x=6
sisi 1 = x = 6
+𝑞
sisi 2 = x + 2 = 6 + 2 = 8 sisi 3 = 2x – 2 = 2 . 6 – 2 = 10
F.
2𝑥−1+1(𝑥+1)
=
𝑥+1 2𝑥−1+𝑥+1 3𝑥
=
𝑥+1
5. A. a = 2 , b = 1 , c = 0
4
D = b2 – 4ac
𝑥
=
𝑥+1
𝑥
4
= 12 – 4 . 2 . 0
𝑥
= 1 > 0 memiliki 2 akar nyata yang
4
berlainan.
X2 + x = 12x
B. a = 1 , b = 2 , c = -3
X2 + x – 12x = 0 X2
D = b2 – 4ac
– 11x = 0
= 22 – 4 . 1 (-3)
a = 1 , b = -11 , c = 0 x1,2 = = X1 = X2 =
= 4 + 12
−(−11)±√(−1)2 −4 .1 .0
= 16 > 0 memiliki 2 akar nyata
2 .1
yang berlainan .
11+11
C. a = 4 , b = -12 , c = 9
2
11+11 2 11−11 2
=
22 2
D = ( -12 )2 – 4 . 4 . 9
= 11
= 144 – 144
0
= 0 = 0 memiliki 2 akar nyata yang
= =0 2
sama D. a = 2 , b = -3 , c = 4
G. a = ( p – 1 )
D = b2 – 4ac
b = 2p
= (-3)2 – 4 . 2 . 4
c=p x1,2 =
=
= = X1 =
= 9 – 32
−2𝑝±√(2𝑝)2 −4(𝑝−1)𝑝
= -23 < 0 tidak memiliki akar nyata
2 .(𝑝−1) −2𝑝±√4𝑝2 −4𝑝2 +4𝑝
6. A. a = 1 , b = m + 1 , c = 9
2 .(𝑝−1)
D=0
−𝑝±2√𝑝
b2 – 4ac = 0
2 .(𝑝−1)
( m + 1 )2 – 4 . 1 . 9 = 0
−𝑝±√𝑝
M2 + 2m + 1 – 36 = 0
𝑝−1
(m+7)(m–5)=0 m = -7 √ m = 5
−𝑝+√𝑝 𝑝−1
Kunci jawaban Matematika kelas x
18
Sagufindo Kinarya
B. a = m – 1 , b = 4 , c = -5
= -11 < 0 sehingga tidak memiliki akar
D=0
nyata ( imajiner )
b2 – 4ac = 0 42 – 4( m – 1 ).(-5) = 0 16 + 20m – 20 = 0
Uji Kompetensi 3.3
20 m = 0 4
m=
20
=
1
1. a = 1 , b = -3 , c = 2
5
−𝑏
A. 𝛼 + 𝛽 = 7. a = p , b = 2 , c = p b2 – 4ac > 0
C.
22 – 4 . p . p > 0 4–
𝑐
B. 𝛼 . 𝛽 =
D>0
4p2
2 1
𝛼+ 𝛽
=
𝛽
=3
= =2
𝑎
1
1
=
𝛼 .𝛽
3 2
= 32 − 2 . 2
( 2p + 2p )(2 – 2p) > 0 √
𝛼
+
−(−3)
D. 𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
>0
P = -1
1
𝑎
=
=9−4
p=1
=5 E. 𝛼 2 𝛽 + 𝛼𝛽 2 = 𝛼𝛽(𝛼 + 𝛽)
8. a = 1 , b = 2p – 1 , c = p2 – 2p + 3
= 2 .3
D<0
=6
– 4ac < 0 ( 2p –1)2 – 4 .1(p2 – 2p + 3) < 0
F.
4p2 – 4p + 1 – 4p2 + 8p – 12 < 0 4p – 11 < 0 4p p
<
𝛽
=
< 11
11
+
𝛽 𝛼
𝛼 2 +𝛽 2
=
𝛼𝛽
5 2
2. a = 3 , b = 5 , c = -2
4
<2
p
𝛼
𝛼+ 𝛽=
D = b2 – 4ac
3 4
𝛼. 𝛽=
= 52 – 4 . 3(-2) 9. a = p , b = - (2p – 3) , c = p + n
+ 4pn = 0 B.
-12p – 4pn + 9 = 0
2 𝛼
+
2 𝛽
=
9 = p ( 12 + 4n ) 12+4𝑛
=p
D=
3
– 4ac
= 32 – 4 . 1 . 5
√49 3
=
7 3
2(𝛼+𝛽) 𝛼𝛽
=
=
=
𝛼𝛽
2 .−
10. a = 1 , b = 3 , c = 5 b2
−2
2𝛼+2𝛽
=
9 = 12p + 4pn 9
√𝐷 𝑎
A. 𝛼 − 𝛽 =
( -(2p – 3))2 – 4 . p ( p + n ) = 0 – 12p + 9 –
=
= 49
b2 – 4ac = 0 4p2
𝑎
𝑎
=
= 25 + 24
D=0
4p2
𝑐
−𝑏
−2 3
−10 3
5 3
×−
3 2
=5
= 9 – 20 Kunci jawaban Matematika kelas x
19
Sagufindo Kinarya
−5 3
3. X2 + 3x – m = 0
1 1 2 12 12𝑥 2 +𝑥−1=0
𝑥 2 + 𝑥− =0
a = 1 , b = 3 , c = -m −𝑏
𝛼+𝛽 =
𝑎
=
−3 1
= −3
× 12
D. 𝛼 + 𝛽 = 2 − √3 + 2 + √3 = 4
𝛼 + 3𝛽 = 5
𝛼 . 𝛽 = (2 − √3)(2 + √3)
𝛼 + 𝛽 = −3
=4−3
2𝛽 = 8
=1
𝛽=4
𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼 . 𝛽 = 0
𝛼 + 𝛽 = −3
𝑥 2 − 4𝑥 + 1 = 0
𝛼 + 4 = −3 𝛼 = −7
-7 . 4 =
3
𝑎
1
7
3
3
=2 =
1
2
3
3
𝛼 .𝛽 = 2 . =
𝑐
𝛼 .𝛽 =
1
E. 𝛼 + 𝛽 = 2 +
𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼 . 𝛽 = 0
−𝑚 1
7
2
𝑥 2 + 3𝑥+ 3= 0
-28 = -m
3𝑥 2 − 7𝑥+2= 0
28 = m
×3
6. A. a = 1 , b = -4 , c = 2 4.
X2
– 6x +2p – 1 = 0
𝛼+𝛽 =
a = 1 , b = -6 , c = 2p – 1 D = b2 – 4ac
𝛼 .𝛽 =
= (-6)2 – 4 . 1 (2p – 1)2
𝑎
−(−4) 1
=4
2
= =2 1
X1 . x2 = 3𝛼 .3𝛽 = 9𝛼𝛽 = 9 .2 = 18
= 40 – 8p
4=
𝑐
𝑎
=
X1 + x2 = 3𝛼 + 3𝛽 = 3(𝛼 + 𝛽) = 3 .4 = 12
= 36 – 8p + 4
X2 – ( x1 + x2 ) x + x1 . x2 = 0
√𝐷 𝑎
𝛼− 𝛽=
−𝑏
X2 – 12
x + 18 = 0
B. X1 + x2 = 3𝛼 − 1 + 𝛽 − 1 = 𝛼 + 𝛽 − 2 =
√40−8𝑝
4−2= 2
1
x1. x2 =3(𝛼 − 1)(𝛽 − 1) = 𝛼𝛽 − 𝛼 − 𝛽 + 1
4 = √40 − 8𝑝
= 𝛼𝛽 − (𝛼 + 𝛽) + 1
16 = 40 – 8p
=2–4+1
8p = 40 – 16
= -1
8p = 24
x2 – ( x1 + x2 ) x + x1 . x2 = 0
P=3
x2 – 2 – 1 = 0
5. A. ( x – 2 )(x – 5 ) = 0 7. a = 1 , b =-2 , c = -3
X2 – 5x – 2x + 10 = 0
𝑥1 + 𝑥2 =
X2 – 7x + 10 = 0 B. ( x + 3 )(x – 7 ) = 0 X2
X1 . x2 =
– 7x + 3x – 21 = 0
𝑐 𝑎
=
X2 – 4x – 21 = 0 1
𝛼+𝛽 =
1
C. ( x – 4 )(x – 3 ) = 0 1
1
−𝑏 𝑎 −3 1
𝑥1 𝑥2
=
−(−2) 1
=2
= −3 𝑥2
+𝑥 = 1
𝑥12 +𝑥
22
𝑥1 𝑥2
1
X2 + 3x – 4x + 12 = 0 Kunci jawaban Matematika kelas x
20
Sagufindo Kinarya
=
= = =
(𝑥1 +𝑥2 )2 −2𝑥1 𝑥2
𝑥 2 − (𝑥1 + 𝑥2 )𝑥 + 𝑥1 . 𝑥2 = 0
𝑥1 𝑥2
7 𝑥 2 − (− ) 𝑥 + (−1) = 0 2 7 𝑥2 + 2 𝑥 − 1 =0 ×2 2𝑥 2 + 7𝑥 − 2 =0
22 −2(−3) (−3) 4 +6
c. 𝑥1 + 𝑥2 = 𝑥 2 + 𝛽+∝ + 𝛽 2
−3 −10 3
= −3
1
=∝3 + 𝛽 3
3
= (∝ +𝛽)3 − 3 ∝ 𝛽(∝ +𝛽) 1 3
1
= (2) − 3. (−2) (2)
2
8. 2𝑥 − 𝑥-4= 0
1 +3 8 25 = 8
a = 2, b = -1, c = -4
=
−𝑏 1 = 𝑎 2 𝑐 −4 ∝. 𝛽 = = = −2 𝑎 2 ∝ +𝛽 =
𝑥1 . 𝑥2 = (∝2 + 𝛽)(∝ +𝛽 2 ) =∝3 +∝2 𝛽 2 +∝ 𝛽 + 𝛽 3
a. 𝑥1 =∝ +1 𝑑𝑎𝑛 𝑥2 = 𝛽 + 1
=∝3 + 𝛽 3 + (∝ 𝛽)2 +∝ 𝛽
𝑥1 + 𝑥2 =∝ +1 + 𝛽 + 1
25 + (−2)2 + (−2) 8 25 = +4−2 8 25 = +2 8 41 = 8
=∝+𝛽+2
=
1
=2+ 2 5
=2 𝑥1 . 𝑥2 = (∝ +1)(𝛽 + 1) =∝ 𝛽+∝ + 𝛽 + 1 = −2 + =−
1 + 1 2
𝑥 2 − (𝑥1 + 𝑥2 )𝑥 + 𝑥1 . 𝑥2 = 0
1 2
25 41 𝑥2 − 8 𝑥 + 8 = 0 ×8 8𝑥 2 − 25𝑥 + 41 = 0
𝑥 2 − (𝑥1 + 𝑥2 ) + 𝑥1 . 𝑥2 = 0 5 1 2 2 2𝑥 2 −5𝑥−1=0
𝑥 2 − 𝑥+(− )=0
×2
9. X2 – x + 2 = 3a + 1 X2 – x – 3a + 2 – 1 = 0 X2 – x – 3a + 1 = 0
b. 𝑥1 =∝ −2 𝑑𝑎𝑛 𝑥2 = 𝛽 − 2
a = 1 , b = -1 , c = -3a + 1
𝑥1 + 𝑥2 = 𝛼 − 2 + 𝛽 − 2 = ∝ +𝛽 − 2
berkebalikan
1 −4 2 7 =− 2 𝑥1 . 𝑥2 = (∝ −2)(𝛽 − 2) =
x1 . x2 = 1 𝑐 𝑎
=1
−3𝑎+1 1
=1
−3𝑎 + 1 = 1
=∝ 𝛽 − 2 ∝ −2𝛽 + 4
1 − 1 = 3𝑎
=∝ 𝛽 − 2(∝ +𝛽) + 4
0 = 3a
1 = (−2) − 2. + 4 2
0=a
= −4 -1 + 4 = −1 Kunci jawaban Matematika kelas x
21
Sagufindo Kinarya
10. a = p – 2
5. x2 – 2x – 8 < 0 ( x – 4 )( x + 2 ) < 0
b = -2 c = 2 + 2a
x=4
misal
x1 = a
√
x = -2
+
0
x2 = 2b ( x – x1)(x – x2 ) = 0
+
√
4
HP = { x | -2 < x < 4 , x ∈ R }
( x – b )( x – 2b ) = 0 6. 3x2 – x2 – 9x + 4 ≤ 0
X2 – 2bx – bx + 2b2 = 0
2x2 – 9x + 4 ≤ 0
X2 – 3bx + 2b2 = 0
( 2x – 1 ) ( x – 4 ) ≤ 0 2x = 1
1.
x=4
1
x=2
Uji Kompetensi 3.4 1 2 𝑥 −2𝑥>0 2 𝑥 2 − 4𝑥>0
√
+
×2
+
1
√
4
2
1
HP = { x | 2 ≤ x ≤ 4 , x ∈ R }
𝑥 (𝑥 − 4) > 0 √
x=0
-
x=4
+
0
√
+
7.
4
3𝑥−2 𝑥
−3<0
3𝑥−2−3𝑥
HP = { x | x < 0 atau x > 4 , x ∈ R }
𝑥 −2
2. ( 2x + 3 ) ( x – 2 ) ≤ 0
𝑥
<0
<0
2x = -3 √ x = 2 x=
−3
+
2
+
−3
√
2
HP = { x |
−3 2
0
+ HP = { x | x > 0 , x ∈ R }
2 8.
≤ 𝑥 0 atau x ≤ 2 , x ∈ R }
𝑥−1 𝑥−2
−
𝑥−3 𝑥−4
≥0
(𝑥−1)(𝑥−4)−(𝑥−3)(𝑥−2)
3.
– 2x – 15 ≥ 0
(𝑥−2)(𝑥−4)
( x – 5 )( x + 3 ) ≥ 0
(𝑥 2 −5𝑥+4)−(𝑥 2 −5𝑥+6)
X2
x=5
√
+
x = -3 √
-3
(𝑥−2)(𝑥−4)
+
𝑥 2 −5𝑥+4−𝑥 2 +5𝑥−6 (𝑥−2)(𝑥−4)
5
HP = { x | x ≤ -3 atau x ≥ 5 , x ∈ R }
−2 (𝑥−2)(−4)
4. 6 + x – x2 < 0
-
√
≥0
≥0
-
x = -2
+ -2
≥0
𝑥 = 2,𝑥 −4
( 3 – x )( 2 + x ) < 0 x=3
≥0
√
√
-
+ 2
4
HP = { x | 2 < 𝑥 < 4 , x ∈ R }
3
HP = { x | x < -2 atau x > 3 , x ∈ R } Kunci jawaban Matematika kelas x
22
Sagufindo Kinarya
5=y L ≥ 136
9. K = 0
HP = { ( 2,5) }
2 ( p + l ) = 50
p x l ≥ 136
p + l = 25
p( 25 – p ) ≥ 136
l = 25 – p
25p – p2 ≥ 136 0≥
p2
3. 5x + y = 2
– 25p + 136
0 ≥ ( p – 8 )(p – 17 ) p = 8 √ p = 17
-
+ 8
5x + y = 5
17x + y = -8
5 (−
-12x
= 10
−
x
=−
-
10
=−
12
25 6
5
y
6
17
7x – 3y = 21
L 2 = s2
X2 + ( 576 – 48x + x2 ) > 20
11
2x2 – 48x + 576 – 20 > 0 2x2 – 48x + 556 > 0
3y =
X2 – 24x + 278 > 0 )
y= =
Uji Kompetensi 3.5
x + 2y = -2
x + 2y = -2
6
4 + 2y = -2
2x
=8
2y = -6
x
=4
y = -3
25
6 1 6
44x
= 84
x
= 44
84
21
11
−
147 11
84 11 11
:3
28
11
=2
6 11
5. 11x + 3y = -7
x5
55x + 15y = -35
2x + 5y = 21
x3
6x +15y = 63 49x x
= -98 = -2
2x + 5y = 21
HP = { ( 4,-3) }
2 (-2) + 5y = 21 2. 3x – y = 1
x2
x – 2y = -8
-4 + 5y = 21
6x – 2y = 2
5y = 21 + 4
x – 2y = -8 5x x
5y = 25
= 10
y=5
=2
HP = { ( -2,5 ) }
x – 2y = -8 2 – 2y = -8 2 + 8 = 2y 10 = 2y Kunci jawaban Matematika kelas x
23
10
= 11 = 1 11
231
84
6
55
35x +15y = 105
+ 3𝑦 = 21 3y =
+
x5
21
147
1. 3x + 2y = 6
30
=
9x – 15y = -21
7 . 11 + 3𝑦 = 21
L1 + L2 > 20
)(x
+y=5
x3
= (24 – x )2 = 576 – 48x + x2
(x
)+y=5
=9 4. 3x – 5y = -7
= x2
6
=
HP = { p | 0 < 𝑝 ≤ 8 atau 17 ≤ p < 25 }
10. L1 = s2
5
Sagufindo Kinarya
8. x – 4 = 0
6. 1 dan 2 2x + 5y + 3z = 6
2x + 5y + 3z= 6
6 x – 6y + 2z = 18 : 2 3x – 3y + z = 9
x=4 x2 – y2 = 16
x3
2x + 5y + 3z = 6
42 – y2 = 16
9x – 9y + 3z = 27
16 – 16 = y2
−7𝑥+14𝑦
=21
−𝑥+2𝑦
= −3
𝑥+𝑦
0 = y2
:7
0=y
=0
3𝑦
HP = { (4,0) }
= −3
y = -1
9. 4x2 – y = 0
x+y=0
4 . 22 – y = 0
x–1=0
4.4
=y
x =1
16 = y
2x + 5y + 3z
=6
HP = { (2,16) }
2 . 1 + 5 (-1) + 3z = 6 2–5
+ 3z = 6
10. x2 + y2 = 7
3z = 9
(√3 )2 + y2 = 7
z =3
3 + y2 = 7
HP = { (1,-1,3) }
Y2 = 4 y = ±√4
7. 1 dan 2 x+y+z=6
= ±2
x 3 3x + 3y + 3z= -3
x + 2y + 3z = 18
HP = { (√3, 2), (√3, −2) }
x + 2y + 3z = -2 2x + y
= -1
1 dan 3 Uji Kompetensi 3.6
x + y + z = -1
1. 3x – 2x < -7
3x – 2y – z = 2 4x – y
x < -7
=1
(B)
3
2
2. 123 . 4 (8𝑥 − 20) + 3,12 ≤ 124 . 3 (6𝑥 + 15) −
4 dan 5 2x + y = -1
4.12
4x – y = 1
9(8𝑥 − 20) + 36
≤ 8(6𝑥 + 15) − 48
6x
=0
72𝑥 − 180 + 36
≤ 48𝑥 + 120 − 48
x
=0
72𝑥 − 48𝑥
≤ 120 = 48 + 180 − 36
2x + y = -1
24𝑥
2 . 0 + y = -1
≤ 216
𝑥
y = -1
3.
x + y + z = -1 0 + (-1) + z = -1
2−3𝑥 >4 5
2−3𝑥 >20
≤9(C)
×5
−18 > 3𝑥
z = -1 + 1
−6 > 𝑥 ( B )
=0 HP = { (0,-1,0) } Kunci jawaban Matematika kelas x
24
Sagufindo Kinarya
2
8. x1 + x2 = 1 – √3 + 1 + √3 = 2
4. 3. 3 (6𝑥 − 12) ≥ 3.2 (6𝑥 + 2) 12𝑥 − 24 ≥ 36𝑥 + 12
x1 . x2 = ( 1 – √3 )( 1 + √3) = 1 − 3 = −2
−24 − 12 ≥ 36𝑥 − 12𝑥
x2 – ( x1 + x2 ) x + x1 . x2 = 0
2
–
2x
2
=0 (B)
9. –
−363 ≥𝑥 242 −3
–
x2
−36 ≥ 24𝑥
3
10. 𝑥1 + 𝑥2 = 1 = 3
≥ 𝑥 (A) 𝑥1 . 𝑥2 =
5. 2 + 1 < 3𝑥 − 1 + 1 < 8 + 1
−4 = −4 1
𝑥12 + 𝑥22 = (𝑥1 + 𝑥2 )2 − 2𝑥1 𝑥2
3 < 3𝑥 < 9 :3 1<𝑥 <3
= 32 − 2(−4) =9+8
6. a = 1 , b = m +1 , c = 2m – 1
= 17
D<0
11. x2 + x – 2 ≥ 0
b2 – 4ac < 0 (m+
1)2
( x + 2 ) ( x -1 ) ≥ 0
– 4 .1( 2m – 1 ) < 0
m2
+ 2m + 1 – 8m + 4 < 0
m2
– 6m + 5 < 0
x = -2 √ x = 1 +
-
+
-2 √ 1
( m – 1 )( m – 5 ) < 0
X ≤ -2 atau x ≥ 1
(A)
m=1√m=5 12. 2c + 3k = 101.500 +
-
√
+
1
c + 2k = 53.500
5
2c + 4k = 107.000
HP = { m | 1 < m < 5 , m ∈ R }
2c + 3k = 101.500 k=
7. a = 2 , b = -1 , c = -5
𝑥1 + 𝑥2 = X1 . x2 =
𝑐 𝑎
=
−𝑏 𝑎
=
2
=
−(−1) 2
=
1
5.500
c + 11.000 = 53.500
2
c
−5
= 42.500
5.500 + 42.500 = 48.000 ( B )
2
+2=
-
c + 25.500 = 53.500
13. x – y = 5
𝛼 + 𝛽 = x1 + 1 + x2 + 1 = x1 + x2 + 2 1
x2
x
5
x=5+y
=5+y
=5+2
x2 – y2 = 45
2
=7
( 5 + y )2 – y2 = 45 25 + 10y + y2 – y2 = 45
𝛼 . 𝛽 = ( x1 + 1 )( x2 + 1 ) = x1x2 + x1 + x2 +1 =−
5 2
1
+ +1
10 y
= 45 – 25
10 y
= 20 y=2
2
HP = { ( 7,2 ) } ( A )
= -1 X2 – ( 𝛼 + 𝛽)𝑥 + 𝛼 . 𝛽 = 0 5
x2 – 2 + ( −1 ) 2𝑥 2 − 5𝑥 − 2 Kunci jawaban Matematika kelas x
=0 =0
×2
(D) 25
Sagufindo Kinarya
14. 4p + 3q =
853.000
x5
3x
= -9
3p + 5q = 1.022.000
x3
x
= -3
20 p + 15 q = 4.265.000
4x + 6y = 4 (-3) + 6(3)
9 p + 15q = 3.066.000
= -12 + 18
11 p
=6
= 1.199.000
p
=
109.000
(A) 18.
2𝑏+2𝑝=8.800 𝑏+𝑝=4.400
15. Ayah = x
:2
b = 600 + p
Anak = y
600 + p + p = 4.400
(x–2)=6(y–2) x–2
(E)
600 + 2p = 4.400
= 6y – 12
2p = 3.800
x = 6y – 10
p = 1900
( x + 18 ) = 2 ( y + 18 )
b = 600 + 1900
x + 18 = 2y + 36
= 2500 ( E )
x = 2y + 18 6y – 10 = 2y + 18 4y
19. 10x + 4y = 31.000
= 28
4x + 10y = 25.000
y
=7
x
= 6y – 10
20x + 8y = 62.000 20x + 50y= 125.000
= 6 . 7 – 10
-
-42y= - 63.000
= 42 – 10
y = 1500 ( A )
= 32
(B) 20. 3b + 2p = 10.500
16. 3x + 4y = 17
x 3 21x + 28y = 119
5x + 7y = 29
x 4 20x + 28y = 116 x
b = p + 1000 3 (p + 1000) + 2p = 10.500
=3
3p + 3000 + 2p = 10.500
3x + 4y = 17
5p
= 7.500
p
= 1.500
b
= 1.500 + 1.000
3 . 3 + 4y = 17 9 + 4y = 17 4y = 8
= 2.500 ( E )
y =2
21. 2x + y
HP = { ( 3,2 ) }
x + 3 z = 16
17. 2x - 5y = -21
2x + y
3x + 2y = -3
x2
=5
2x + 6z = 32
6x – 15y = -63 6x + 4y = - 6
=5
-
y – 6z = -27 -
5y - z
x5
= 10
-19y = -57 y=3
5y – 30z = -135
3x + 2.3 = -3 3x
5y - z
=-3–6
Kunci jawaban Matematika kelas x
26
=
10 -
29z = -145 Sagufindo Kinarya
26. x2 + 5x – m = 0
z=5 5y – 5 = 10
a = 9, b = 5, c = -m
5y
= 15
𝛼 .𝛽 = 3
y
=3
𝑐 𝑎
x + 3.5 = 16 x
=3
−𝑚 9
=1
=3
−𝑚 = 27
{(1, 3, 5)}
𝑚 = −27
(D)
27. akar – akar kembar
22. 6x2 – 11x – 10 < 0
D=0 b2 – 4ac = 0
( 2x – 5 ) ( 3x + 2 ) < 0 2x = 5 √ 3x = -2 x=
5
=0
144 – 16a
=0
−2
x=
2
( -12 )2 – 4.a.4
3
144 = 16a 9 =a
+
−2
√
3
{x|
−2 3
<x<
5
9x2 – 12x + 4 = 0
+ 5
( 3x – 2 )2
2
3x – 2 = 0 3x = 0
,x∈R}
2
=0
2
x=3
23. D
(C)
28. A
24. X2 + x – 12 < 0
29. 3x2 – 4x + 1 = 0
(x+4)(x–3)<0
( 3x – 1 ) ( x – 1 ) = 0
x = -4 √ x = 3
3x = 1 √ x= 1 +
-4
1
x=3
+ √
3
30. Akar khayal
(E) D<0 b2 – 4ac < 0
{ x | -4 < x < 3 , x ∈ R }
52 – 4(p – 2 ).(-2) < 0
{ -3, -2, -1, 0, 1, 2 }
25 + 8( p – 2 ) < 0
25. a = 1, b = 2 , c = -a
25 + 8p – 16 < 0
D=
b2
– 4ac
=
22
– 4 . 1 ( -a )
8p
< -9 p<
= 4 + 4a
−9 8
(A)
31. ( x – 7 )( x + 3 ) = 0
X1 – x2 = 8
X2 + 3x – 7x – 21 = 0
√𝐷 =8 𝑎
X2 – 4x – 21 = 0
√4+4𝑎 =8 1
32. ( x – √2 )( x – √8 ) = 0 X2 – √8 x – √2 x + √16 = 0
√4 + 4𝑎 = 8
X2 – 3 √2 x + 4 = 0
4 + 4𝑎 = 64
(E)
4a = 60 a = 15
(A)
Kunci jawaban Matematika kelas x
27
Sagufindo Kinarya
33. X2 + x – 6 = 0
A. Baris = 5
( x + 3 )( x – 2 ) = 0
Kolom = 4
X1 = -3 √ x2 = 2
B. 5, 7, 7, 5
𝛼 = -3 + 2 𝛽 = 2 + 2
C. 7, 7, 8, 6, 8
= -1
D. 4
=4
1 2 2. A = [5 6 9 1
(x+1)(x–4)=0 X2 – 3x – 4 = 0 ( A )
3 4 1 5 9 7 8] transposenya [2 6 1 ] 2 3 3 7 2 4
B = [1
2 3 4 5 6]
34. A 𝑐
Berkebalikan
=1
𝑎
−𝑏
35. Berlawanan
𝑎
=0 C = [ [1 2
36. a = b, b = c, c = -a x1,2 =
= =
−𝑏±√𝑏2 −4𝑎𝑐 2𝑎
1 2 3 4 5 6 1 2 3 4 5
−𝑐±√𝑐 2 −4𝑏(−𝑎) 2.𝑏
1 3
−𝑐±√𝑐 2 −4𝑎𝑏
(D)
2𝑏
D=
−𝑏
x1 + x2 = X1 . x2 =
𝑎 𝑐
=
𝑎
−𝑏
=
−𝑎
𝑐 −𝑎
=
=
−𝑐 𝑎
(E)
−𝑏
= 9x2 + 12x − 36
𝑎
=
∶3
3. – (A)
3 1 −2 4 6 ] + 3[ ]=[ −2 0 5 −3 −4 −6 12 [ ] 15 −9 0 14 =[ ] 11 −9 3 1 −2 4 3 B. ([ ]+[ ]) − ([ −2 0 5 −3 −2 2 1 [ ]) 1 0 1 5 5 2 =[ ]−[ ] 3 −3 −1 0 −4 3 = [ ] 4 −3 2 1 3 1 C. 3 [ ] −2 [ ] 1 0 −2 0
4. A. 2 [
3 1
1
(E)
40. E
Uji Kompetensi 4 . 1 5 7 7 5 6 8 7 3 8 7 8 2 6 4 6 1 7 6 8 0 Kunci jawaban Matematika kelas x
5 7 9 ] 6 8 1
7 2 4 3 3 −1 1. [ ]+[ ]=[ ] 4 −5 2 −1 2 −4 2𝑥 8 4𝑥 −10 6𝑥 −2 2. [ ]+[ ]=[ ] 3𝑥 −7 2𝑥 12 5𝑥 5
−1
( x1 + x2 )2 = ( − 3 )2 = 9
1 3 2 4
Uji Kompetensi 4 . 2
= 9x2 + x2 – x2 + 12x – 36
0 =3𝑥 2 +4𝑥−12
39. X1 + x2 =
[
𝑎
36 – 12x + x2 = 9x2 + x2 0
5 6
𝑏
38. ( 6 – x )2 = (3x)2 + x2
0
2 4
7 8 9 1
37. a = -a, b = b, c = c
1. P =
3 4 5]
8 3
28
2 ]+ 0
1 ]+ 0
Sagufindo Kinarya
6 3 0 = [ 7
3 6 2 ]− [ ] 0 −4 0 1 ] 0 4 2 −2 3. [−1] − [ 7 ] = [−8] 3 −6 9 4. –
2+0 6+0 2 0 1 3 ][ ]=[ ]= 1 + 6 3 + 12 1 3 2 4 2 6 [ ] 7 15 2 0 −2 4 BC= [ ][ ]= 1 3 1 5 −4 + 0 8 + 0 −4 8 [ ]= [ ] −2 + 3 4 + 15 1 19
4 −2 ] 3 1 4 2 B = [−3 −1] 0 −5 1 −7 2 −3 6. A. A = [ ]− [ ] 8 0 4 −1 2 −4 = [ ] 4 1 −2 −4 2 3 B. A = [ ]+ [ ] 7 5 1 4 0 −1 = [ ] 8 9
1 2 B. At = [ ] 3 4 2 1 Bt = [ ] 0 3 −2 1 Ct = [ ] 4 5 5 9 −2 4 C. ( AB )C = [ ][ ]= 8 12 1 5 −10 + 9 20 + 45 −1 65 [ ]= [ ] −16 + 12 32 + 60 −4 92 1 3 −4 8 D. A ( BC ) = [ ][ ]= 2 4 1 19 −4 + 3 8 + 57 −1 65 [ ]= [ ] −8 + 4 16 + 76 −4 92
= [
BA =[
5. A = [
7. 3 c = 4b C=
E. AB – BC = [5
4𝑏
8
−4 8 9 9 ]−[ ]=[ 1 19 7 12
1 ] −7
3
8. X + 2 – 4 = 3
F. A ( B – C ) = [
X–2=3 X
=3+2
7+y+1=y 9. 2x + 3 + 2 = 7 2x = 2 x =1 3+y+2=5 y =0 x+y=1+0=1 −10 3 −7 8 ]−[ ] 12 5 4 −6 −18 ] 18 −9 ] 9
I. AtBt = ( BA )t BtAt = ( AB )t 3 24 15 2. A. [ ] [8 5] = [ ] 4 32 20 4 3 2 −3 1 B. [ ] [−2 1] = 4 0 2 0 2 8+6+0 6−3+2 [ ] 16 + 0 + 0 12 + 0 + 4
Uji Kompetensi 4 . 3 1 3 2 ][ 2 4 1 5 9 [ ] 8 12
1. A. AB = [
2+3 0 ]=[ 4+4 13
Kunci jawaban Matematika kelas x
0 −2 4 ]−[ ]) 3 1 5
1 3 4 −4 =[ ][ ] 2 4 0 −2 4 + 0 −4 − 6 4 −10 =[ ]=[ ] 8 + 0 −8 − 8 8 −16 5 8 G. ( AB )t = [ ] 9 12 2+0 1+6 1 2 2 1 H. AtBt = ( )( )=[ ]= 6 + 0 3 + 12 3 4 0 3 2 7 [ ] 6 15 2+3 4+4 2 1 1 2 BtAt = [ ][ ]=[ ]= 0 + 9 0 + 12 0 3 3 4 5 8 [ ] 9 12
=5
9 3 10. 2p = [ 7 6 6 10 =[ 2 2 3 5 P=[ 1 1
1 3 2 ] ([ 2 4 1
0+9 ]= 0 + 12
29
Sagufindo Kinarya
B. D = 3 . 2 – 1 . 6 = 6 – 6 = 0
14 5 =[ ] 16 16 px + qy p q x 3. A = [ ] [ ] = [ rx + sy ] r s y 2 B= [ 5
3 −1 0 3 C. |1 2 3| 1 4 3 2 4
2x + 3y 3 x ] [y] = [ ] 5x + 7y 7
4. 2p – 6q = -16
D = ( 12 – 12 + 0 ) – ( 0 – 27 – 2 ) = 29
p – 3q = -8
x5
2 1 0 2 1 D. |−1 3 2| −1 3 4 2 5 4 2
-5p + 8q = -13 -5p + 8q = -13 3x + 3y + 3z= -3
D = ( 30 + 8 – 0 ) – ( 0 + 8 – 5 )
x + 2y + 3z = 18
= 38 – 3
-7q = -53 q=
= 35
53 7
p – 3q = -8 p–3. p-
159 7
53 7
2. A. = -8 −56
=
B.
7
p=
10 2.10−6.3 −3
0 2 ][ 1 5 3 1 ][ 7 0
3 2 ]=[ 7 5 0 2 ]=[ 1 5
3 ] 7 3 ] 7
C. D.
B. Ya E.
1 10 −6 ]= [ 2 −3 2
[
1
−2 2(−2)−5(−1) −5
[
5
−6 ]= 2
−3
[− 3
1
2
1 1 −2 ] = −6+5 [ 3 −5
]
1 ]= 3
−1 ] −3
[
7
1 0 2 AI=[ 5
1
2 5
103
5. A. I A = [
−1 2 3
1
3 2.3−1.5 −1
1 3 −5 ] = 1[ 2 −1
[
1
3 [ 5.3−2.7 −7
1 3 −2 ] = 1[ 5 −7
1
3 [ 3.3−5.2 −2
−5 ]= 3
1 −1
[
−5 3 ]=[ 2 −1 −2 3 ]=[ 5 −7
3 −2
−5 ] 2 −2 ] 5
−5 −3 ]=[ 3 2
5 ] −3
3. A. x ( x – 5 ) – 2 . 3 = 0
6. A.
X2 – 5x – 6 = 0
B. Tidak
(x–6)(x+1)=0 x = 6 √ x = -1
1+3 1+2 1 1 1 1 ][ ]=[ ]= 3+6 3+4 3 2 3 2 4 3 [ ] 9 7 4+9 4+6 4 3 1 1 P3 = [ ][ ]=[ ]= 9 + 21 9 + 14 9 7 3 2 13 10 [ ] 30 23 −1 2 −1 2 8. A2 = [ ][ ]= 1 3 1 3 1 + 2 −2 + 6 3 4 [ ]=[ ] −1 + 3 2 + 9 2 11 7. P2 = [
1 3 −1 1 3 B. | 3 8 2 | 3 8 −1 x −3 −1 x ( -24 + ( -6 ) – 3x ) – ( 8 + 2x – 27 ) = 0 -30 – 3x + 19 – 2x -11 = 5x − 3.5−2.7
=
−1 2 4 ] − 2[ ]−5=0 1 3 11 −2 4 4 ]−[ ]−5 =0 2 6 11 0 ]−5 =0 5
[
[
5 −7
11 5
−2 ]= 3
=𝑥 1
15−14
5 −7
[
−2 ] 3
5 −2 ] −7 +3
B. AA1=[3 2] [−2 1] = [15 − 14 7 5 −5 3
1 =[ 0 C. A1A=[ 5
−7
−6 + 6 ] 35 − 35 −14 + 15
0 ] 1 −2 3 ][ 3 7
2 15 − 14 ]=[ 5 −21 + 21
10 − 10 ] −14 + 15
1 0 =[ ] 0 1
Uji Kompetensi 4 . 4
D. AA-1 = A-1A = I
1. A. D = 4 ( -2 ) – 3 ( -2 ) = -8 + 6 = -2 Kunci jawaban Matematika kelas x
1
4. A. A-1 =
A2 – 2A – 5 = 0 3 [ 2 3 [ 2 5 [ 0
=0
30
Sagufindo Kinarya
3 1 9 5. A. [ ]A = [ ] 3 2 12 A=
2 6−3 −3 1
[
3 −1 −2 4 ][ ] 6 0 −4 2 1 −6 − 6 12 − 0 x = 2[ ] 8 + 12 −16 + 0 1 −12 12 x = 2[ ] 20 −16 −6 6 x=[ ] 10 −8 −4 6 2 3 B. [ ]x = [ ] −7 8 1 4 1 4 −3 −4 6 x = 8−3 [ ][ ] −1 2 −7 8 1 −16 + 21 24 − 24 x = 5[ ] 4 − 14 −6 + 16 1 0 x=[ ] −2 2 5 2 6 5 C. x [ ]=[ ] 3 1 1 0 6 5 1 0 −5 x =[ ] [ ] 1 0 0−5 −1 6 0 − 5 −30 + 30 1 =[ ] 0−0 −5 + 0 −5 1 0 =[ ] 0 1 2 1 −2 4 D. x [ ]=[ ] 8 5 0 6 −2 4 1 5 −1 x =[ ] 10−8 [ ] 0 6 −8 2 −10 − 32 2 + 8 1 =[ ] 0 − 48 0 + 12 2 −21 5 =[ ] −24 6 1
x = 6−4 [
−1 9 ][ ] 3 12
18 − 12 ] −27 + 36 1 6 = 3[ ] 9 2 =[ ] 3 1 4 6 18 B. A – [ ]=[ ] 2 5 1 7 6 18 1 4 A=[ ]+[ ] 2 5 1 7 7 22 =[ ] 3 12 −8 2 −5 x 6. A. [ ] [y] = [ ] −13 3 −8 x 1 −8 5 −8 [y] = −16+15 [ ][ ] −3 2 −13 1 64 − 65 = −1 [ ] 24 − 26 x 1 [y] = [ ] 2 1 2 x 11 B. [ ] [y] = [ ] 2 −1 2 x 1 −1 −2 11 [y] = −1−4 [ ][ ] −2 1 2 1 −11 − 4 = −5 [ ] −22 + 2 1 −15 = −5 [ ] −20 x 3 [y] = [ ] 4 1 2 5 1 2 7. A. |1 x =0 5 |1 x 3 −1 −2 3 −1 1
=3[
9. – −2 1 0 −2 1 10. | 1 −2 1| 1 −2 2 −3 0 2 −3 D=(0+2–0)–(0+6+0)
( -2x + 30 – 5 ) – ( 15x – 5 – 4 ) = 0
=2–6
-2x + 25 – 15x + 9 = 0 34
= -4
=0
−2 1 ]=0+3=3 −3 0 1 1 A12 = [ ] = −(0 − 2) = 2 2 0 1 −2 A13 = [ ] = −3 + 4 = 1 2 −3 1 0 A21 = [ ] = −(0 + 0) = 0 −3 0 −2 0 A22 = [ ]=0+0=0 2 0 −2 1 A23 = [ ] = −(6 − 2) = −4 2 −3 1 0 A31 = [ ]=1+0=1 −2 1 A11 = [
2=x 1 2 B. |1 x 1 4
−3 1 2 −3| 1 x x 1 4
=0
( x2 – 6 – 12 ) – ( -3x – 12 + 2x ) = 0 X2 – 18 + 3x + 12 – 2x = 0 X2 + x – 6 = 0 (x+3)(x–2)=0 x = -3 √ x = 2 2 8. A. [ 4
1 −2 4 ]x = [ ] 3 6 0
Kunci jawaban Matematika kelas x
31
Sagufindo Kinarya
−2 0 ] = −(−2 − 0) = 2 1 1 −2 1 A33 = [ ]= 4−1= 3 1 −2 3 0 1 [2 0 2] 𝐴𝑑𝑗 𝐴 A-1 = = 1 −4 3 |𝐴| −4 A32 = [
− = −
3 4 1 2 1
0 − 0 −
7. (
1 5 5 −2 −2 5 −2 [ ] = 1[ ]=[ ] −7 3 −7 3 −7 3 1 4 −6 8 9. 𝑥 = 20−18 [ ][ ] −3 5 6 1 32 − 36 = 2[ ] −24 + 30 1 −4 = 2[ ] 6 −2 =[ ] (B) 3
8.
1 4 1 2 3
[− 4 1 − 4]
x = A-1B
1. a = 2
1 −2 −2 5 2 = 2−6 [ ][ ] −3 −1 −3 −6 1 −10 + 6 −4 + 12 = −4 [ ] −15 + 3 −6 + 6 1 −4 8 =−4 [ ] −12 0 1 −2 =[ ] (D) 3 0 0 1 0 1 11. A2 = [ ][ ] −2 3 −2 3 0−2 0+3 =[ ] 0 − 6 −2 + 9 −2 3 =[ ] (C) −6 7 2 3 1 2 3 12. |0 −1 2| 0 −1 2 4 1 2 4
b = 2a = 2 . 2 = 4 (D)
−2 3 −7 −5 1 2. A + B = [ ]+[ ]=[ −1 5 3 4 −3
4 ] (A) 2
3. –a = 4 a = -4 2a = 3a = 3 (-4) = -12 b = -6 c = -b = - ( -12 ) = 12 x = 2c = 2 . 12 = 24 5x = 5 . 24 = 120 4. 2p – 1 = 11 2p p
= 12 =6
1
15−14
10. Ax = B
Uji Kompetensi 4 . 5
c = 2b = 2 . 4 = 8
−1 − 6 −3 − 6 5 )=[ ] −2 + 8 −6 + 8 −3 4 −19 =( )(B) −7 17
−2 3 1 )( 1 −4 2
2q + 3 = -9 2q
= -12
q
= -6
= ( -2 + 12 + 0 ) – ( -2 + 16 + 0 ) = 10 – 14 = -4 ( B )
2r + 1 = 5 2r
=4
r
=2
13. Matrik singular
D=0 -15x – 30 = 0 -30 = 15x
p + q + r = 6 + (-6) + 2 = 2 ( C )
-2 = x
2 −1 12 4 5 8 )−( ) + 2( ) 3 7 −8 9 −6 2 −3 −9 24 8 =( )+( ) 9 5 −16 18 21 −1 =( ) (A) −7 23 −7 −1 −2 6 11 −13 6. ( )+( )−( ) −5 −10 −9 12 −4 3 −20 18 =( ) (B) −10 −1 5. (
2 −4 ] −6 8 1 2 −4 = [ ] −8 −6 8
14. A-1 =
1
16−24
[
1
1
− 4 =[ 3
2
−1
4
x
15. [y] = Kunci jawaban Matematika kelas x
(E)
32
1 6−16
[
]
(D)
6 4 −3 ][ ] 4 1 2
Sagufindo Kinarya
−18 + 8 ] −10 −12 + 2 1 −10 = [ ] −10 −10 =
1
x 1 [y] = [ ] 1
x
a=4
[
6b = a + 8 6b = 4 + 8 6b = 12
(E)
b=2
3 −2 2 3 p ][ ][ ] 1 1 5 −2 q 6 − 10 9 + 4 p =[ ][ ] 2+5 3−2 q −4 13 p =[ ][ ] ( C ) 7 1 q
2c = 3a – b
16. [y] = [
17. A-1
=3.4–2 2c = 10 c=5 −1
4 2 2 1 21. 2 [ 2 ] + 3 [0] + k [1] = [−3] 1 3 3 −2 −2
Bt
=
−2 1 𝑎 −𝑏 [ ]=[ ] −8−3 −3 4 𝑥 𝑐 1 −2 1 𝑎 −𝑏 [ ]=[ ] −5 −3 4 𝑥 𝑐 1
2
[53 5
b=
− −
−2 12 2 2 [ 1 ] + [ 0 ] + k [1] = [−3] +1 9 3 −2 10 2 2 [ 1 ] + k [1] = [−3] +10 3 −2 10 2 2 k [1] = [−3] − [ 1 ] +10 3 −2 2 −8 k [1] = [ −4 ] 3 −18
1 5 4]
=[
5
1
𝑎 𝑥
−𝑏 ] 𝑐
(B)
5
k = -4
18. Det A = Det B
−1 − 6 −3 − 6 1 −3 −1 −3 22. [ ][ ]=[ ] −2 + 8 −6 + 8 2 4 2 2 −7 −9 =[ ] (B) 6 2 1 2 1 2 3 23. [ ] [3 4] = 4 5 6 5 6 1 + 6 + 15 2 + 8 + 18 22 28 [ ]=[ ] 49 64 4 + 15 + 30 8 + 20 + 36
2x2 – 9 = 10x + 3 2x2 – 10x – 9 – 3 = 0 2x2 – 10x – 12 = 0 X2 – 5x – 6 = 0 (x–6)(x+1)=0 x = 6 √ x = -1
(A)
3 4 1 3 4 19. |2 x 5| 2 x 3 2 3 3 2
(A) 1 4 −2 24. 𝐴−1 = 4−6 [ ] −3 1 −1 4 −2 = 2 [ ] (A) −3 1
Det = 0 ( 6x + 60 + 4 ) – ( 3x + 30 + 16 ) = 0 ( 6x + 64 ) – ( 3x + 46 ) = 0
=
25. 3+ 4y = 11
6x + 64 – 3x – 46 = 0
4y = 8
3x
y=2
= -18 x
20. 2A
(E)
= -6 ( E )
5+z+y=4
Bt
5+z+2=4
2a + 2 a + 4 5 a ]=[ ] a + 8 3a − b 3b c 2𝑎 + 2 𝑎 + 4 10 2a [ ]=[ ] 𝑎 + 8 3𝑎 − 𝑏 6b 2c
z
2[
3x – z + 2
= 14
3x – (-3) + 2 = 14 3x
2a = a + 4 Kunci jawaban Matematika kelas x
= -3
33
=9 Sagufindo Kinarya
x
=3
C. S
x + y + z = 3 + 2 + (-3) = 2 ( D )
D. B E. B
Uji Kompetensi 6 . 1 Uji Kompetensi 6 . 2 1. A. Pernyataan Benar
1. A. 1. 2 + 3 = 5
B. Bukan Pernyataan
2. 5 adalah bilangan ganjil
C. Pernyataan Benar
B. 1. Hari ini mendung
D. Pernyataan
2. Hari ini gerah
E. Pernyataan Salah
C. 1. 4 x 5 = 20
2. A. Ada siswa SMK tidak rajin belajar
2. 20 bilangan prima
B. Dua bilangan prima
D. 1. Mas Boi mencintai Iteung
C. Semua murid yang menganggap
2. Iteung tidak mencintai Mas Boi
matematika itu mudah.
E. 1. 3 bukan bentuk akar
D. Ada demonstran tidak memakai ikat
2. 3 = √3
kepala. E. Kuadrat setiap bilangan real tidak selalu
2. A. B ∧ S = S
tak negative.
B. B ∧ B = B
3. A. S
C. S ∧ S = S
B. S
D. B ∧ B = B
C. S
E. B ∧ S = S
D. B
3. A. Saya ingin belajar dan naik kelas
E. B 4. A.
B.
x2
B. Saya ingin belajar tetapi tidak naik kelas + 7x + 10 = 0
C. Saya tidak ingin belajar tetapi naik kelas
(x+5)(x+2)=0
D. Saya tidak ingin belajar atau naik kelas
x = -5 √ x = -2
E. Saya tidak ingin belajar dan naik kelas
2log
x=4
2x
= 11+ 3
= 16
2x
= 14
x= C. 32x-1
=
4. A. 2x – 3 = 11
24
33
x
2x – 1 = 3 2x
=4
x
=2
=7
B. X2 – 4x + 4 = 0 ( x – 2 )( x – 2 ) = 0 x=2√x=2
D. 3x + 7 = 31
C. 2 + x = 10
3x = 24
x=8
x=8
D. 3log x = 2
E. 4x – 12 ≤ 0
x = 32
≤0
=9
4x x
≤3
5. p q p ∧ q ∼p ∼q ∼( p∧ q ) ∼p∧∼q
5. A. B
B B
B
S
S
S
S
B. S
B S
S
S
B
B
S
Kunci jawaban Matematika kelas x
34
Sagufindo Kinarya
S B
S
B
S
B
S
D. p
q
r
∼q
∼q∧r
S S
S
B
B
B
B
B
B
B
S
S
B
B
B
S
S
S
B
6. A. S V B = B
B
S
B
B
B
B
B. B V B = B
B
S
S
B
S
B
C. B V S = B
S
B
B
S
S
S
D. S V S = S
S
B
S
S
S
S
S
S
B
B
B
B
S
S
S
B
S
S
7. A. 2x – 3 = 11 2x
= 14
p∧ (∼q v r )
9. A. ( p ∧ q ) v r
x=7 B. X2 – 4x + 4 = 0
B. ( p v r ) ∧ ( q v s )
( x – 2 )( x – 2 ) = 0
C. ( q v r ) v p
x=2
8. A. p
q
r
∼q
p∧∼q
B
B
B
S
S
B
B
B
S
S
S
S
B
S
B
B
B
B
B
S
S
B
B
B
S
B
B
S
S
B
S
B
S
S
S
S
S
S
B
B
S
B
S
S
S
B
S
S
B. p
q
r
∼q
∼q∧r
B
B
B
S
S
S
B
B
S
S
S
S
B
S
B
B
B
B
B
S
S
B
S
S
S
B
B
S
S
S
S
B
S
S
S
S
S
S
B
B
B
S
B. ~p
q
S
S
S
B
S
S
C. ~p
~q
C. p
q
r ∼p ∼q ∼pv∼q (∼pv∼q )∧ r
B
B
B
S
S
S
B
B
S
S
S
B
S
B
S
B
S
S
S
B
S
(p∧∼q) v r
10. A.
q
P
B.
P
𝑞 r
p
p∧ (∼q v r )
r
C.
q
s
Uji Kompetensi 6 . 3 1. A. p
q
2. A. S
B=B
S
B. S
S=B
S
S
C. B
B=B
B
B
B
D. B
B=B
S
B
B
S
E. S
B=B
B
B
S
B
B
F. S
S=B
B
S
B
S
B
S
S
S
B
B
B
B
B
3. A. p
S
S
S
B
B
B
S
B
Kunci jawaban Matematika kelas x
r
35
~p
p v ~p
S
B
Tautology Sagufindo Kinarya
S
B
B
B B B
S
S
B
B. p
q
pΛq p
B B S
S
S
B
B
B
B
B
B S B
B
B
B
B
S
S
S
B S S
B
B
S
S
B
S
B
S B B
S
S
B
S
S
S
B
S B S
S
S
B
C. p
q
pvq p
S S B
B
S
B
B
B
B
B
S S S
B
S
B
B
S
B
B
S
B
B
B
B B B S S
B
S
S
S
S
S
B
B B S S B
B
B
B
B S B S S
S
S
B
B S S S B
S
S
B
pvq
B. p q r ∼p ∼r ∼pvq q∧∼r ∼pvq
Taulogi
q
B
B
B
B
B
S B B B S
B
S
S
B
S
B
S
S
S B S B B
B
B
B
S
B
B
S
S
S S B B S
B
S
S
S
S
S
B
B
S S S B B
B
S
S
q ~(p
p
q (p v q)
(p
q) ~p ~q ~q
q)
q∧∼r
4. p
5. p q
pvq p
pΛq
~p
9. p q ∼p ∼q p
q ∼pv∼q ~q
~p p
B B
B
S
S
S
B
B B S
S
B
B
B
B
B S
S
B
S
B
S
B S S
B
S
S
S
S
S B
B
S
B
S
B
S B B
S
B
B
B
S
S S
B
S
B
B
B
S S B
B
B
B
B
B
~p
~q
p ∧ ∼q
p
q∧q
B
S
B
B
B
S
S
S
S
B
S
B
6. A. ~q
~p
B. p Λ ~q
p
A. p
q ≡ ∼p v ∼q
B. p
q ≡ ~q
C. p
q ≡ (p
~p q) ∧ (q
p)
7. A. Sx = 25 x=5
10. A. Ahmad tidak merokok atau ia batuk
B. X3 = 8
Jika Ahmad tidak batuk maka ia tidak
X=2
merokok
C. X – 1 = 3x – 9
B. Hujan tidak turun atau halaman basah
-1 + 9 = 3x – x
Jika halaman tidak basah maka hujan
8 = 2x
tidak turun
4 =x Uji Kompetensi 6 . 4 8. A. p q
r ∼q p∧∼q
Kunci jawaban Matematika kelas x
(p∧∼q)
r 36
Sagufindo Kinarya
q
1. A. Konvers = Jika Warman lari pagi maka hari libur
q ~(p
p ∧∼q
q
= jika hari tidak libur maka
B
B
B
S
S
S
Warman tidak lari pagi.
B
S
S
B
B
B
Kontraposisi = Jika Warman tidak lari
S
B
B
S
S
S
pagi maka hari tidak libur .
S
S
B
S
B
S
Invers
p
q) ~ q
3. p
Kesimpulan : ~( p
B. Konvers = Jika Dulkamdi lulus tes maka
q ) ≡ p ∧ ∼q
ia belajar. Invers
= Jika Dulkamdi tidak belajar
4. A. Besi bukan logam atau tidak keras
maka ia tidak lulus tes.
B. Kotak tidak dapat hidup di darat dan di
Kontraposisi = Jika Dulkamdi tidak lulus
darat
tes maka ia tidak belajar.
C. Guru tidak datang dan murid tidak
C. Konvers = Jika Rosi disenangi oleh guru
senang
maka ia berwajah cantik. Invers
D. Tuti tidak cantik atau tidak pemalu
= Jika Rosi tidak berwajah cantik
E. Tuti manis atau tidak cantik
maka ia tidak disenangi oleh
F. Ia tidak di rumah dan tidak masuk
guru.
sekolah
Kontraposisi = Jika Rosi tidak disenangi
G. Hujan turun dan air tidak meluap
oleh guru maka ia tidak
H. Ia tidak naik kelas dan ia diberi hadiah
berwajah cantik.
I. Ia tidak sehat jika dan hanya jika dia tertawa – tertawa
D. Konvers = Jika Rosiman seorang anggota DPR maka ia seorang anggota MPR. Invers
= Jika Rosiman bukan seorang
Uji Kompetensi 6 . 5
anggota MPR maka ia bukan anggota
1. A. p
DPR.
~q
Kontraposisi = Jika Rosiman bukan
∴~p
seorang anggota DPR maka ia
Ia tidak sakit
bukan seorang anggota MPR. 2. A. Konvers : ~q
∴~p ~p
Amir tidak belajar
p
C. p
Invers : ~(p Λ q)
~r
Kontraposisi : ~r
~(p Λ q)
C. Konvers : ~(~q Λ r) Invers : ~p
Invers : ~( p
p
uang
(p
q)
q)
r
Kunci jawaban Matematika kelas x
r Jika Ali tamat SMK mak ia dapat
~p
~(p
r
∴p
(~q Λ r)
Kontraposisi : r
q
q
Kontraposisi : (~q Λ r) D. Konvers : ~r
q
~q
q
Kontraposisi : q B. Konvers : r
B. p
p
Invers : ~p
q
2. A. Valid B. p
q
~p
tidak Valid
∴~q
q) 37
Sagufindo Kinarya
10. Ada orang tidak berdiri ketika tamu agung C. p
q
memasuki ruangan.
q
tidak Valid
11. Jika sungai tidak banyak ikan maka sungai
∴P 3. p
tidak dalam ( E )
q
p
12. ~( p ∧ ∼p )
r
∼p
( ∼p v q )
q
~r
~r
∴ ~𝑝
∼p
(C)
13. Jika ia seorang eropa maka ia seorang
Ia tidak rajin belajar
Belanda .
(E)
14. p ∧ ~(p Vr)
4. p
q
q
r
p
r
r
s
r
s
p
p∧~q∧~r p
p
s
~P
~q ∧ ~ r ( D )
p
15. Jika x ≤ 2 maka 2x + 1 ≤ 5
∴s
16. Air laut tenang dna nelayan tidak melaut
Kelelawar berkaki dua
mencari ikan
5. A.
q ≡ ∼q
17. p
B. p v q q
~p
r
∴
(D)
p
r
C. p v q
r
p
r
∴p
r
Tidak Valid
19. p ∧ ∼q Perang terjadi dan ada orang tidak gelisah .
q
(E)
p
∴q
(B)
18. A
q
~p
∼p
20. –
Tidak Valid
∴q
21. B
D. –
22. C 23. E 24. p
q
Uji Kompetensi Materi Pokok 06
q
r
p
r
1. D
r
s
r
s
2.
q
∼p
B
B
S
S
B
S
B
B
S
S
B
S
S
B
S
B
B
S
B
S
B
S
S
B
B
B
B
B
p
∼q p q p
q ∧∼q {(p q) ∧∼q}
∼p
∴p
Jika Dinda rajin belajar maka ia bahagia . (B) 25. p
5.
∴∼ p ( B )
[ p v ( q Λ r ) ] Λ [ (~p Λ ~q ) v ~r ] p
q
p
q
∼q
3. C atau E 4.
s
q ∼q (p q) ∧∼q
{(p
q) ∧∼q}
B B
B
S
S
B
B S
S
B
S
B
S B B S
S
B
S S B B
B
B
(C) ∼p
26. p ∧ ∼q Nafila tidak rajin belajar dan ada temannya senang. ( D )
6. B 7. x2 = 25 dan x ≠ 5 (D) 8. E
27. p
q
∼q
q ∼q ∧ p ( p
B
B
S
B
S
B
S
B
S
B
p
q)v
9. E Kunci jawaban Matematika kelas x
38
Sagufindo Kinarya
S
B
S
B
S
S
S
B
B
S
∴p
r (A)
46. Ada peserta ujian nasional yang tidak lulus
( ∼q ∧ p )
(B)
B
47. p
q
B
∼q
B
∴∼ p ( B )
B
48. Jika hasil karya manusia tidak baik maka
28. p ∧ q , q B
S S
r , dan r
s
B
S
B
sumber daya manusia tidak baik. ( D ) 49. ∼q
( D ) atau ( E ) q ≡ ∼p v q
29. p
(D)
31. p
50. p
q
q
r
∴p
30. Saya lulus ujian tetapi saya tidak kuliah
∼p ( A )
r (A)
q
∼q ∴∼ p ( B ) 32. E 33. D 34. p
q
∼q ∴ ∼p
(C)
35. Jika ia diduga tidak bersalah maka ia bukan tersangka ( B ) 36. E 37. 3x + 2 = -1 3x = -3 x = -1
(A)
38. – 39. p
q
∼q ∴ ∼p 40.. p r p
(D) q
p
q
∼q
q
∼r
p
p
∼r
p ∴ ∼p
(A)
41. E 42. D 43. ∼(p ∧ q )=∼ p ∧∼ q (B) 44. B 45. p
q
q
r
Kunci jawaban Matematika kelas x
39
Sagufindo Kinarya
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