Question

帮帮我吧···要过程 初二数学题 速度

xy²/(x²y-y) × x²/(x²+x)=
(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=

化简求植
x²-1/(x²-x-2)除以x/2x-4,其中x=1/2
[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1

Answer 1

xy²/(x²y-y) × x²/(x²+x)=xy/(x²-1 ) × x/(x+1 ) =x²y/[(x+1)² × (x-1)]

(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=x-3/x-5 × 2(x-5)/(x-3)² =2/(x-3)

x²-1/(x²-x-2)除以x/2x-4,=(x+1)(x-1)/(x-2)(x+1) × 2(x-2)/x=2(x-1)/x,代入x=1/2得-2

[x-x/(x+1)]除以[x/(2x-4)] =x²/x+1 × 2(x-2)/x=2x(x-2)/(x+1)其中x=√2+1,得2-√2