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Soal-soal Eksponen 1. Bentuk sederhana dari ( 1 + 3 2 ) – ( 4 –
50 ) adalah ….
a. – 2 2 – 3 b. – 2 2 + 5 c. 8 2 – 3 d. 8 2 + 3 e. 8 2 + 5 Jawab (1+3 2)–(4–
50 ) = ( 1 + 3 2 ) – ( 4 –
25.2 )
=(1+3 2)–(4– 5 2 )=1+3 2–4+ 5 2 =–3+ 8 2
7x 2. Nilai dari
(1 + 2 (1 + 2 (1 + 2 (1 + 2 (1 + 2
a. b. c. d. e.
−.
3 2 6
y5
1 −. 54 x − 6 y 3 x −2 2 ).9 2 2 ).9 3 2 ).18 3 2 ).27 2 2 ).27 3
untuk x = 4 dan y = 27 adalah ….
Jawab 7x
−.
3 2 6
x − 6y 5 4
=
=
=
y5 −2 x
1 −. 3
7(4)
−.
3 2
=
.(27)
7x
5 2
52 2 − 6.3 −1 2 − 4 7 .3 2 . 3
(2
)
2 −1
x
5 6
3 2
.y . 1 −. x − 6 y 3 x −2 5 4
5 6
7( 2 2 )
1 −. 54 (4) − 6(27) 3 (4) − 2
7.2 −.3.3
−.
−.
3 2
5
.(33 ) 6 = 1 −. 2 54 (2 ) − 6(3 3 ) 3 (2 2 ) − 2 2+
1
7.2 −.3.3 2 .2 4 7.2.3 2 . 3 7.2.3 2 . 3 = = = 2 + 12 2 2 . 2 − 2 2 2. 2 − 1 1 2 − 6. 3
2 2 +1 2 2 +1
(
=
) (
7.9 3 (2 2 + 1) = 9 3 (2 2 + 1) (8 − 1)
)
50 ) adalah ….
a. – 2 2 – 3 b. – 2 2 + 5 c. 8 2 – 3 d. 8 2 + 3 e. 8 2 + 5 Jawab (1+3 2)–(4–
50 ) = ( 1 + 3 2 ) – ( 4 –
25.2 )
=(1+3 2)–(4– 5 2 )=1+3 2–4+ 5 2 =–3+ 8 2
7x 2. Nilai dari
(1 + 2 (1 + 2 (1 + 2 (1 + 2 (1 + 2
a. b. c. d. e.
−.
3 2 6
y5
1 −. 54 x − 6 y 3 x −2 2 ).9 2 2 ).9 3 2 ).18 3 2 ).27 2 2 ).27 3
untuk x = 4 dan y = 27 adalah ….
Jawab 7x
−.
3 2 6
x − 6y 5 4
=
=
=
y5 −2 x
1 −. 3
7(4)
−.
3 2
=
.(27)
7x
5 2
52 2 − 6.3 −1 2 − 4 7 .3 2 . 3
(2
)
2 −1
x
5 6
3 2
.y . 1 −. x − 6 y 3 x −2 5 4
5 6
7( 2 2 )
1 −. 54 (4) − 6(27) 3 (4) − 2
7.2 −.3.3
−.
−.
3 2
5
.(33 ) 6 = 1 −. 2 54 (2 ) − 6(3 3 ) 3 (2 2 ) − 2 2+
1
7.2 −.3.3 2 .2 4 7.2.3 2 . 3 7.2.3 2 . 3 = = = 2 + 12 2 2 . 2 − 2 2 2. 2 − 1 1 2 − 6. 3
2 2 +1 2 2 +1
(
=
) (
7.9 3 (2 2 + 1) = 9 3 (2 2 + 1) (8 − 1)
)
