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第84页

信息发布者：

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

$=2-\frac {x+2y}{3x+y}$

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

$=\frac {6x+2y-x-2y}{3x+y}$

$=\frac {5x}{3x+y}$

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

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

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

$=\frac 1{x-3}$

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

$=\frac {y(x-y)}x-\frac {xy}{x}$

$=-\frac {y^2}x$

$=(\frac {x^2-4x+3}{x-3}+\frac 1{x-3})· [\frac {(x-1)^2}{(x-1)(x-2)}-\frac 2{x-2}]$

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

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

$=x-2$

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

当$x=3$时，$A=\frac {3-1}{3+1}=\frac 12$

解：原式$=[\frac 3{a+1}-\frac {(a-1)(a+1)}{a+1}] · \frac {a+1}{(a-2)^2}= \frac {3-a^2+1}{a+1}\ \mathrm {·} \frac {a+1}{(a-2)^2} =\frac {4-a^2}{(a-2)^2}=\frac {(2-a)(2+a)}{(2-a)^2}=\frac {a+2}{2-a}$

由分式有意义的条件可知$a≠-1，$$a≠2$

∴故$a$可取$a=0，$ ∴原式$=\frac 22=1$

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

$ $由$x^2+2x-3=0，$解得$x_{1}=-3，$$x_{2}=1$

∵$x≠1，$ ∴当$x=-3$时，原式$=-\frac 1{2×(-3)-1}=\frac 17$