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第89頁(yè)

信息發(fā)布者：

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}$

$當(dāng)a=-2，b=3時(shí)， $

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

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第89頁(yè)