第81页 - 经纶学典新课时作业八年级数学上下册【江苏国标】

0或-1

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{6}{5}\ $

$解:原式=\frac{y(x-2)}{(x-2)²}=\frac {y}{x-2}$

$∵x=1，y=-2，∴原式=2.$

$解:原式=\frac {(x+y+x-y)(x+y-x+y)}{xy(x+y)}=\frac{4xy}{xy(x+y)}=\frac{4}{x+y}，\ $

$∵x=3，y=1，∴原式=1.$

$解:原式=\frac{a(a+2)²}{a(a+2)(a-2)}=\frac{a+2}{a-2}，$

$a只有取±1时，原分式才有意义.$

$若取a=-1，原式=-\frac{1}{3};若取a=1，原式=-3.$

$ \begin{aligned} 解:\frac{a^3+b^3}{a^3+(a-b)^3}&=\frac {(a+b)(a^2-ab+b^2)}{[a+(a-b)][a^2-a(a-b)+(a-b)^2)]} \\ &=\frac { (a+b)(a^2-ab+b^2)}{[a+(a-b)](a^2-a^2+ab+a^2-2ab+b^2)} \\ &=\frac {(a+b)(a^2-ab+b^2)}{[a+(a-b)](a^2-ab+b^2)} \\ &=\frac {a+b}{a+(a-b)} \\ \end{aligned}$