第89页 - 经纶学典新课时作业八年级数学上下册【江苏国标】

a+1

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

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

（更多请点击查看作业精灵详解）

$ \begin{aligned}解:原式&=\frac{2x}{(x+y)(x-y)}·\frac{(x+y)(x-y)}{x²y} \\ &=\frac {2}{xy} \\ \end{aligned}$

$∵y=\frac{2}{x}，∴xy=2，∴原式=1.$

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

$∵a²-a-1=0，即a²=a+1，∴原式=1.$

$ \begin{aligned}解:原式&=(x-\frac {1}{x})(x+\frac {1}{x})(x^2+\frac {1}{x^2}) (x^4+\frac {1}{x^4}) (x^8+\frac {1}{x^8})·(x^{16}+\frac {1}{x^{16}})(x^2-1)÷(x-\frac {1}{x}) \\ &=(x^2-\frac {1}{x^2}) (x^2+\frac {1}{x^2}) (x^4+\frac {1}{x^4}) (x^8+\frac {1}{x^8})(x^{16}+\frac {1}{x^{16}})·(x^2-1)÷(x-\frac {1}{x}) \\ &=(x^{32}-\frac {1}{x^{32}})x \\ &=x^{33}-\frac {1}{x^{31}} \\ &=\frac {x^{64}-1}{x^{31}} \\ \end{aligned}$

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

$因为当a=1，2时分式无意义，$

$所以a=3. $

$当a=3时，原式=\frac{3}{2}$