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第7页

信息发布者：

$=\frac {x-1}{x} · \frac {x^2}{x-1}$

$=x$

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

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

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

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

$=\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} $

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

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

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

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

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

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

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

$ \begin{aligned}解：(1)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} \\ \end{aligned}$

$当x=-1时，原式=\frac {3}{-1-1}=- \frac {3}{2}$

$ \begin{aligned}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)} \\ \end{aligned}$

$由a^2+3a+2=0，得到a^2+3a=-2，即a(a+3)=-2$

$则原式=- \frac {1}{6}$