Aljebra Ggzzi Sol 08-09

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EBAZPENA 1) LABURTU: x 5 + 6x 4 + 9x 3 x 3 + 3x 2 Ebazpena:

(

)

x 3 x 2 + 6x + 9 x 5 + 6x 4 + 9x 3 x 3 ( x + 3) = = 2 = x ( x + 3) = x 2 + 3 x 3 2 2 x + 3x x ( x + 3) x ( x + 3) 2

2) Egin eta laburtu:

( x − 1) 2 2



1 3x − x − 1 ( x + 1) 2 2

Ebazpena:

( x − 1) 2 2 =

1 3x ( x − 1) 3x − = − = 2 2 2 ( x − 1) ( x + 1) ( x + 1) 2 x − 1 ( x + 1) 2



x −1 3x x 2 − 1 − 6x x 2 − 6x − 1 − = = 2 2 2 ( x + 1) ( x + 1) 2 2 ( x + 1) 2 ( x + 1)

3) Ebatzi ondoko ekuazioak: a) x ( x + 4 ) − 5 =

x ( x − 1) 3

b) x 4 − 48 x 2 − 49 = 0

Ebazpena: x ( x − 1) 3 2 x −x x 2 + 4x − 5 = 3 3 x 2 + 12 x − 15 = x 2 − x

a) x ( x + 4 ) − 5 =

2 x 2 + 13 x − 15 = 0 x=

− 13 ± 169 + 120 4

=

− 13 ±

289

4

=

x = 1 − 13 ± 17  →  − 30 − 15 4  x = 4 = 2

=

z = 49 → x = ±7 48 ± 50  →  2 z = −1 (no vale)

b) x 4 − 48 x 2 − 49 = 0 Cambio: x 2 = z



x 4 = z2

z 2 − 48 z − 49 = 0 z=

48 ±

2 304 + 196 2

=

48 ±

Bi soluzio: x1 = −7, x2 = 7 4)Ebatzi ondoko ekuazioak:

2 500 2

a)

x +5 − x =3

b)

4x x 14 + = x +2 x −2 3

Ebazpena: a)

x +5 −x =3 x +5 =3+ x x + 5 = 9 + x 2 + 6x 0 = x 2 + 5x + 4 x=

−5±

25 − 16 − 5 ± 9 − 5 ± 3 = = 2 2 2



 x = −1   x = −4

Egiaztapena: x = −1 →

4 +1= 2 +1= 3

x = −4

1 + 4 = 1+ 4 = 5 ≠ 3





x = −1 sí vale →

x = −4 no vale

Soluzioa: x = −1 b)

4x x 14 + = x +2 x −2 3 14( x + 2) ( x − 2) 12 x ( x − 2) 3 x ( x + 2) + = 3( x + 2 ) ( x − 2 ) 3( x + 2 ) ( x − 2 ) 3( x + 2) ( x − 2)

(

12 x 2 − 24 x + 3 x 2 + 6 x = 14 x 2 − 4 2

)

2

15 x − 18 x = 14 x − 56 x 2 − 18 x + 56 = 0 x=

18 ±

324 − 224 18 ± 100 18 ± 10 = = 2 2 2

c) x 4 + x 3 − 4x 2 − 4x = 0 Faktorizatu:

(

)

x 4 + x 3 − 4x 2 − 4x = x x 3 + x 2 − 4x − 4 = 0 Factorizamos x 3 + x 2 − 4 x − 4 :



 x = 14  x = 4

x 4 + x 3 − 4 x 2 − 4 x = x ( x + 1) ( x − 2) ( x + 2) = 0



x = 0  x + 1 = 0 → x = −1  x − 2 = 0 → x = 2 x + 2 = 0 → x = −2

Soluzioak: x 1 = 0,

x 2 = −1,

x 3 = 2,

x 4 = −2

5) Ebatzi ondorengo ekuazioak: a)

2 4 x −1 2 3 x +2

b) log x 2 + log 4 = −2

= 16

Ebazpena: a)

2 4 x −1 23 x +2

= 16

2 4 x −1−( 3 x +2 ) = 16



2 4 x −1−3 x −2 = 2 4



2 x −3 = 2 4

→ x −3 = 4



x =7

Soluzioa: x = 7 b) log x 2 + log 4 = −2 log (4x2) = −2 1 100 1 1 1 x2 = → x=± =± 400 400 20 1 1 Hay dos soluciones: x1 = − ; x 2 = 20 20 4 x 2 = 10 −2

4x 2 =



6) Ebatzi ondoko ekuazio sistemak a) 3 x  − = 0 x y  2 x − y = 3 Ebazpena: 3 x  − =0 x y  2 x − y = 3 

3y − x 2 = 0   2 x − y = 3 

0 = x 2 − 6 x + 9;

x=



x2 3 x2 2x − = 3; 3 y=

36 − 36 2

=

6x − x 2 = 9

6 =3 2



y =3

Soluzioa: x = 3; y = 3 b)  y2 − x = 2  log ( x + y ) = 1 Ebazpena:  y2 − x = 2  log ( x + y ) = 1

y + y − 12 = 0 2

y2 −2 = x

(

)

log y 2 − 2 + y = 1 →



y=

−1 ±

y 2 − 2 + y = 10

1 + 48 −1 ± 49 −1 ± 7 = = 2 2 2

− y =3 → x =9−2 =7 − y = −4 → x = 16 − 2 = 14 Bi soluzio: x1 = 7, y1 = 3 x2 = 14, y2 = −4

7)Ebatzi ondoko inekuazio edota inekuazio-sistemak: 3x − 2 < 4   2 x + 6 > x − 1 Ebazpena: 3x − 2 < 4   2 x + 6 > x − 1

3 x < 6  x > −7

x<2 x > −7

Soluzioa: {x < 2 y x > −7} = {x / −7 < x < 2} = (−7, 2)



y = 3    y = −4 

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