第90页 - 经纶学典新课时作业八年级数学上下册【江苏国标】

D

B

C

mc

x

x-1

$\frac{1}{x-y}$

1

$ \begin{aligned}解:原式&=\frac {(x+y)^2}{y(x-y)}×\frac {x(x-y)}{y(x+y)}·\frac {y}{x-y} \\ &=\frac {x(x+y)}{y(x-y)} \\ \end{aligned}$

（更多请点击查看作业精灵详解）

$ \begin{aligned} 解:原式&=\frac{x}{x-y}·y \\ &=\frac{xy}{x-y}， \\ \end{aligned}$

$∵(x-y-4)²+|xy+3|=0，$

$∴x-y=4，xy=-3，$

$∴原式=-\frac{3}{4}$

（更多请点击查看作业精灵详解）

$ \begin{aligned}解:原式&=\frac {3}{2+a}+\frac {1}{2-a}-\frac {2a}{(2+a)(2-a)} \\ &=\frac {3(2-a)}{(2+a)(2-a)}+\frac {2+a}{(2+a)(2-a)}-\frac {2a}{(2+a)(2-a)} \\ &=\frac {6-3a+2+a-2a}{(2+a)(2-a)} \\ &=\frac {8-4a}{(2+a)(2-a)} \\ &=\frac {4(2-a)}{(2+a)(2-a)} \\ &=\frac{4}{2+a} \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac {(a-1)(a-3)+1}{a-3}×\frac {a-3}{a^2-4} \\ &=\frac {a^2-4a+3+1}{a-3}×\frac {a-3}{(a+2)(a-2)} \\ &=\frac {a^2-4a+4}{a-3}×\frac {a-3}{(a+2)(a-2)} \\ &=\frac {(a-2)^2}{a-3}×\frac {a-3}{(a+2)(a-2)} \\ &=\frac{a-2}{a+2} \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac {xy^2}{(x+y)(x-y)}-\frac {x^3y}{(x^2+y^2)(x^2-y^2)}×\frac {x^2+y^2}{x} \\ &=\frac {xy^2}{(x+y)(x-y)}-\frac {x^2y}{(x+y)(x-y)} \\ &=\frac {xy^2-x^2y}{(x+y)(x-y)} \\ &=\frac {xy(y-x)}{(x+y)(x-y)} \\ &=-\frac {xy}{x+y} \\ \end{aligned}$