RESPUESTAS AL ALGEBRA DE BALDOR

Page 17

5.

nx + cn − ab 8nx − 2cn − ab

10.

− ab − 5

nx

− 6 + x 4 − x 2y2

5x 3 y

10nx − cn − 3ab − 5 1 1 ⇒ 10 ⋅ ⋅ 5 − 10 ⋅ − 3 ⋅ 2 ⋅ 3 − 5 5 5 50 10 = − − 6 ⋅3 − 5 5 5 = 10 − 2 − 18 − 5 = − 15 6.

x 3 y − xy 3 + 5 − 6xy + 2 + 3xy 3 + 1

− y4

6x 3y − 4xy 3 + 2 + x 4

− y4

⇒ x + 6x y − 4xy − y + 2 4

3

3

4

= 5 4 + 6 ⋅ 53 ⋅ 4 − 4 ⋅ 5 ⋅ 43 − 44 + 2 = 625 + 24 ⋅125 − 20 ⋅ 64 − 256 + 2 = 625 + 3.000 − 1280 − 256 + 2 = 2.091 .

+ b3

a3

+ x2 y2

3

− 3a b + 8ab − b 3 2

2

− 5a 3

− 6ab 2 + 3a b

− 4a

+8 − 2b

2

11.

3

2

⇒ − 4 ⋅2 + 2⋅2⋅3 − 2⋅3 + 8 = − 4 ⋅ 8 + 4 ⋅ 9 − 2 ⋅ 27 + 8 = − 32 + 36 − 54 + 8 = − 42 3

2

3

+

3 2 − 2 +1 6 + 1− 3 2 a − ab + b 4 6 9 3 1 4 = a 2 − ab + b 2 4 6 9 4 1 3 ⇒ ⋅ 22 − ⋅ 2 ⋅ 3 + ⋅ 32 9 6 4 3 4 1 = ⋅ 4 − ⋅ 6 + ⋅ 9 = 3 − 1+ 4 = 6 6 4 9

+ 125n 3

3 7. 27m

− 9m n + 25mn 2

2

− 14mn 2

−8

+ 10m n + 11mn 2 2

27m3 + m2 n + 22mn 2 + 125n 3 − 8 3

2

2

3

2  1  2  2 1  1 ⇒ 27 ⋅   +   ⋅ + 22 ⋅ ⋅   + 125 ⋅   − 8  3  3 5  5 3  5 8 4 1 44 1 1 = 27 ⋅ + ⋅ + ⋅ + 125 ⋅ −8 27 9 5 3 25 125 4 44 4 44 =8+ + + 1− 8= + +1 45 75 45 75 20 + 132 + 225 377 = = 1 152 = 225 225 225 a −1 + y b − 2 + mx − 4 8. x

2x

− 2y

b−2

x −4

− 2m

+ 3y

b−2

− 2m x − 4

3x a −1 + 2 y b − 2 − 3mx − 4  2 ⇒ 3⋅ 52 −1 + 2 ⋅ 4 3 − 2 − 3   3 9.

n

b −1

− m

x −3

+8

− 5nb −1 − 3m x − 3 + 10 4nb −1 + 5m x − 3 − 18 mx − 3  2 ⇒   3

5−3

4  2 =  =  3 9 2

5−4

= 15 + 8 − 2 = 21

2 2 b 3 1 1 ab + b 2 3 9 1 1 ab − b 2 6 3 +

+ 2ab − 2b3 + 8

3

a −1

3 2 a 4

12.

9 2 m 17 7 2 m 34 7 2 − m 34

+ +

25 2 n 34 5 2 n 17

− + − +

1 4 1 2 1 4 3

− 15mn − 30mn

18 + 7 − 7 2 25 + 10 2 − 1 + 2 − 1 + 12 − 45mn m + n + 34 34 4 18 2 35 2 12 = − 45mn m + n + 34 34 4 35 2 9 2 ⇒ m − 45mn + n +3 34 17 2 2 2 1 35  1 9  2 =   − 45⋅ ⋅ +   +3 3 5 34  5 17  3  9 4 90 35 1 ⋅ − + ⋅ +3 17 9 15 34 25 4 7 35 4 = − 6+ + 3= + −3 17 170 850 17 40 + 7 − 510 463 = =− = − 2 123 170 170 170 =


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