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第99页

信息发布者：

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解：原式$=\frac {1+\mathrm {m^2}-2m}{m-1}$

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

$=m-1$

解：原式$=\frac {x+3}{(x-1)^2}·\frac {x-1}{x(x+3)}+\frac {1}{x}$

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

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

$=\frac {x}{x(x-1)}$

$=\frac {1}{x-1}$

解：原式$=\frac {x+2-1}{x+2}÷\frac {x+1}{(x+2)^2}$

$=\frac {x+1}{x+2}·\frac {(x+2)^2}{x+1}$

$=x+2$

解：原式$=(\frac {3a}{(a+1)(a-1)}-\frac {a+1}{(a+1)(a-1)})÷\frac {2a-1}{a+1}$

$=\frac {3a-(a+1)}{(a+1)(a-1)}·\frac {a+1}{2a-1}$

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

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

解：原式$=(\frac {1}{x+1}+\frac {x+1}{x+1})÷\frac {(x+2)(x-2)}{(x+1)^2}$

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

$=\frac {x+1}{x-2}$

$ $当$x=-4$时，

$ $原式$=\frac {-3}{-6}=\frac {1}{2}$

$ (1)$证明：

$ \begin {aligned}f(\mathrm {x})+f(\dfrac {1}{x})&=\dfrac {2x}{x+1}+\dfrac {2·\dfrac {1}{x}}{\dfrac {1}{x}+1}\\&=\dfrac {2x}{x+1}+\dfrac {\dfrac {2}{x}}{\dfrac {1+x}{x}}\\&=\dfrac {2x}{x+1}+\dfrac {2}{x+1}\\&=\dfrac {2x+2}{x+1}\\&=\dfrac {2(x+1)}{x+1}\\&=2\end {aligned}$

$ (2)$解：

$ $由$(1)$知$f(\mathrm {x})+f(\dfrac {1}{x})=2，$

$ $则$f(\dfrac {1}{101})+f(101)=2，$$f(\dfrac {1}{100})+f(100)=2，$$\dots ，$$f(\dfrac {1}{2})+f(2)=2，$

$ $共有$100$对这样的组合，又$f(1)=\dfrac {2×1}{1+1}=1，$

$ $所以原式$=2×100+1=201$