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第7页

信息发布者：

$=\frac {x-1}x· \frac {x^2}{x-1}$

$=x$

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

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

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

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

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

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

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

$=[\frac {a-2}{a+2}-\frac {a+8}{(a+2)^2}]· \frac {a+2}{a-4}$

$=\frac {(a+2)(a-2)-(a+8)}{(a+2)^2}· \frac {a+2}{a-4}$

$=\frac {a^2-a-12}{a+2}· \frac 1{a-4}$

$=\frac {(a-4)(a+3)}{a+2}· \frac 1{a-4}$

$=\frac {a+3}{a+2} $

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

$=\frac {3x-6}{(x-1)^2}· \frac {x-1}{x-2}$

$=\frac {3(x-2)(x-1)}{(x-1)^2(x-2)}$

$=\frac 3{x-1}$

当$x=-1$时，原式$=\frac 3{-1-1}=-\frac 32$

解：原式$=\frac {a-3}{3a(a-2)}÷\frac {(a+2)(a-2)-5}{a-2}=\frac {a-3}{3a(a-2)}· \frac {a-2}{(a+3)(a-3)}=\frac 1{3a(a+3)}$

由$a^2+3a+2=0，$得$a^2+3a=-2，$即$a(a+3)=-2$

则原式$=-\frac 16$