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第95页

信息发布者：

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

$\frac{2m - 2}{m^2}$

解：原式$= \frac {3y(x+3)\ \mathrm {·} 2(x-3)}{(x-3)\ \mathrm {·} 9y^2} $

$= \frac {6y(x+3)(x-3)}{9y^2(x-3)} $

$= \frac {2(x+3)}{3y}$

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

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

$= -(x + y) $

$= -x - y$

解：原式$= \frac {xy - y^2 - x^2 - xy}{(x - y)(x + y)}\ \mathrm {·} \frac {x + y}{x^2 + y^2}$

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

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

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

 解：$\because \sqrt{x - 3} + y^2 - 4y + 4 = 0，$即$\sqrt{x - 3} + (y - 2)^2 = 0$

 $\therefore x - 3 = 0，$$y - 2 = 0，$

解得$x=3，$$y=2$

      $\frac{x^2 - y^2}{xy} · \frac{1}{x^2 - 2xy + y^2} ÷ \frac{x}{x^2 y - xy^2}$

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

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

 $= \frac{x + y}{x}$

 当$x=3,y=2$时，原式$= \frac{3 + 2}{3} = \frac{5}{3}$