第89页 - 经纶学典学霸八年级数学上下册【江苏国标】

B

C

ab

a+1

1

-4

$ \begin{aligned} 解:原式&=\frac{a²}{(a-1)²}·\frac{a-1}{a}-\frac{1}{a-1} \\ &=\frac{a}{a-1}-\frac{1}{a-1} \\ &=\frac{a-1}{a-1} \\ &=1. \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac{3m+n}{m}·\frac{m}{(3m+n)(3m-n)} \\ &= \frac{1}{3m-n}. \\ \end{aligned}$

$ \begin{aligned}原式&= \frac{a+2b}{a+b}- \frac{a-b}{a-2b} ·\frac{(a-2b)^{2} }{(a+b)(a-b)} \\ &= \frac{a+2b}{a+b} -\frac {a-2b}{a+b} \\ &= \frac{4b}{a+b}. \\ \end{aligned}$

$ \begin{aligned} 解:原式&=[\frac {(a-1)(a+ 2)} {a+2}+\frac {a+3}{a+2}]·\frac{a+2}{(a+1)(a-1)} \\ &=\frac{a²-a+2a-2+a+3}{a+2}· \frac{a+2}{(a+1)(a-1)} \\ &= \frac{a²+2a+1}{(a+1)(a-1)} \\ &=\frac{(a+1)²}{(a+1)(a-1)}=\frac{a+1}{a-1}. \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac{a-b}{a²}÷\frac {a²+b²-2ab}{a} \\ &=\frac {a-b}{a²}·\frac{a}{(a-b)²} \\ &=\frac{1}{a(a-b)}， \\ \end{aligned}$

$当a=-2，b=3时， $

$原式=\frac{1}{-2×(-2-3)} =\frac{1}{10}.$