第71页 - 江苏版实验班提优训练数学八年级上下册答案

A

$\frac{b-2}{b²-9}\ $

$ \begin{aligned}解:原式&=\frac{4}{2x} \\ &=\frac{2}{x}. \\ \end{aligned}$

$解:原式=\frac{a+1}{a-1}-\frac{4a}{(a+1)(a-1)}\ $

$=\frac{(a+1)(a+1)}{(a-1)(a+1)}-\frac{4a}{(a+1)(a-1)}\ $

$=\frac{a²+2a+1-4a}{(a+1)(a-1)}\ $

$=\frac{(a-1)^{2} }{(a+1)(a-1)}$

$=\frac {a-1}{a+1}.$

$ \begin{aligned}解:原式&=\frac{(x-1)^{2} }{(x-1)(x+1)}+\frac{x^{2} }{x+1}\ \\ &=\frac{x-1}{x+1}+\frac{x^{2} }{x+1} \\ &=\frac{x²+x-1}{x+1}.\ \\ \end{aligned}$

$∵x²-2=0，$

$∴x²=2.$

$∴原式=\frac{2+x-1}{x+1}=\frac{x+1}{x+1}=1.$

因式分解

三

$解:∵\frac{Ax+B}{x+1}+\frac{C}{x-2}$

$=\frac{(Ax+B)(x-2)+C(x+1)}{(x+1)(x-2)}\ $

$=\frac{Ax²+(B+C-2A)x+C-2B}{(x+1)(x-2)}$

$=\frac{3}{(x+1)(x-2)},\ $

$∴\begin{cases}{ A=0, }\ \\ {B+C-2A=0,\ }\\{C-2B=3,} \end{cases}\ $

$解得\begin{cases}{ A=0, }\ \\ { B=-1, } \\{C=1.}\end{cases}\ $

$解:\frac{2x}{x²-4}-\frac{1}{x+2}$

$ \begin{aligned}&=\frac{2x}{(x+2)(x-2)}-\frac{1}{x+2} \\ &=\frac{2x}{(x+2)(x-2)}-\frac{x-2}{(x+2)(x-2)} \\ &=\frac {2x-x+2}{ (x+2)(x-2)} \\ &= \frac {x+2} {(x+2)(x-2)} \\ &=\frac{1}{x-2}. \\ \end{aligned}$