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第71页

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$-x - 2$

$ \begin{aligned}解:原式&=\frac {x+y-y-2x+y}{y-x} \\ &=1 \\ \end{aligned}$

解：原式$=\frac {(a+2)(a-2)}{a+2}+2$

$=a-2+2$

$=a$

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

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

 解：$\because ab = 1，$

 $\therefore M=\frac{1}{1 + a}+\frac{1}{1 + b}=\frac{1 + b + 1 + a}{1 + a + b + ab}=\frac{a + b + 2}{a + b + 2}=1，$

 $N=\frac{a}{1 + a}+\frac{b}{1 + b}=\frac{a + ab + b + ab}{1 + a + b + ab}=\frac{a + b + 2}{a + b + 2}=1，$

 $\therefore M = N.$

 解：$\because f(n)+f(\frac{1}{n})=\frac{2n}{n + 1}+\frac{2\cdot\frac{1}{n}}{\frac{1}{n}+1}=\frac{2n}{n + 1}+\frac{2}{n + 1}=\frac{2n + 2}{n + 1}=2，$

 $\therefore$原式$=f(1)+[f(2)+f(\frac{1}{2})]+[f(3)+f(\frac{1}{3})]+\cdots+[f(100)+f(\frac{1}{100})]+[f(101)+f(\frac{1}{101})]=1 + 2\times100=201.$