Question

ajuda !!!! pff
se os números x e y são tais que....

Answer 1

Vamos lá

x³ - 4x² + 4x = x*(x² - 4x + 4) = x*(x - 2)²

x³ - 6x² + 12x - 8 = (x - 2)³

x*(x - 2)² /(x - 2)³ = x/(x - 2) (d)

Answer 2

x³-4x²+4x =x*(x²-4x+4) =x*(x-2)²

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x³-6x²+12x-8= x³-8 -6x*(x-2)

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(x-2)³=x³-3*x²*2 +3 *x*2²-2³

(x-2)³=x³-2³-3x*2(x-2)

x³-2³=(x-2)³+3*x*2(x-2)

x³-8 =(x-2)³+6x*(x-2)

x³-8 =(x-2)*[(x-2)²+6x]

x³-8 =(x-2)*(x²+2x+4]

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Ficamos com:

x³-6x²+12x-8= (x-2)*(x²+2x+4) -6x*(x-2)

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y=(x³-4x²+4x)/(x³-6x²+12x-8)

y=x*(x-2)²/[(x-2)*(x²+2x+4) -6x*(x-2)]

y=x*(x-2)/[(x²+2x+4) -6x]

y=x*(x-2)/(x² - 4x + 4)  

.....sabemos que (x²-4x+4)=(x-2)²

y=x*(x-2)/(x-2)²

y=x/(x-2)

Letra D