登录
1. 计算：
(1) $ 2 - \frac{3x + y}{x - 2y} ÷ \frac{9x^{2} + 6xy + y^{2}}{x^{2} - 4y^{2}} $；
(2) $ \frac{x + 2}{x^{2} - 9} ÷ (1 - \frac{1}{x + 3}) $；
(3) $ (x - \frac{y^{2}}{x}) · \frac{y}{x + y} - y $；
(4) $ (\frac{x^{2} - 4x + 3}{x - 3} - \frac{1}{3 - x})(\frac{x^{2} - 2x + 1}{x^{2} - 3x + 2} - \frac{2}{x - 2}) $。
(1) $ 2 - \frac{3x + y}{x - 2y} ÷ \frac{9x^{2} + 6xy + y^{2}}{x^{2} - 4y^{2}} $；
(2) $ \frac{x + 2}{x^{2} - 9} ÷ (1 - \frac{1}{x + 3}) $；
(3) $ (x - \frac{y^{2}}{x}) · \frac{y}{x + y} - y $；
(4) $ (\frac{x^{2} - 4x + 3}{x - 3} - \frac{1}{3 - x})(\frac{x^{2} - 2x + 1}{x^{2} - 3x + 2} - \frac{2}{x - 2}) $。
答案
解：原式$=2-\frac {3x+y}{x-2y}·\frac {(x+2y)(x-2y)}{(3x+y)^2}$
$=2-\frac {x+2y}{3x+y}$
$=\frac {2(3x+y)-(x+2y)}{3x+y}$
$=\frac {6x+2y-x-2y}{3x+y}$
$=\frac {5x}{3x+y}$
解：原式$=\frac {x+2}{(x+3)(x-3)}÷\frac {x+3-1}{x+3}$
$=\frac {x+2}{(x+3)(x-3)}÷\frac {x+2}{x+3}$
$=\frac {x+2}{(x+3)(x-3)}·\frac {x+3}{x+2}$
$=\frac {1}{x-3}$
解：原式$=\frac {(x+y)(x-y)}{x}·\frac {y}{x+y}-y$
$=\frac {y(x-y)}{x}-\frac {xy}{x}$
$=-\frac {y^2}{x}$
解：原式$=(\frac {x^2-4x+3}{x-3}+\frac {1}{x-3})·[\frac {(x-1)^2}{(x-1)(x-2)}-\frac {2}{x-2}]$
$=\frac {(x-2)^2}{x-3}·(\frac {x-1}{x-2}-\frac {2}{x-2})$
$=\frac {(x-2)^2}{x-3}·\frac {x-3}{x-2}$
=x-2
$=2-\frac {x+2y}{3x+y}$
$=\frac {2(3x+y)-(x+2y)}{3x+y}$
$=\frac {6x+2y-x-2y}{3x+y}$
$=\frac {5x}{3x+y}$
解：原式$=\frac {x+2}{(x+3)(x-3)}÷\frac {x+3-1}{x+3}$
$=\frac {x+2}{(x+3)(x-3)}÷\frac {x+2}{x+3}$
$=\frac {x+2}{(x+3)(x-3)}·\frac {x+3}{x+2}$
$=\frac {1}{x-3}$
解：原式$=\frac {(x+y)(x-y)}{x}·\frac {y}{x+y}-y$
$=\frac {y(x-y)}{x}-\frac {xy}{x}$
$=-\frac {y^2}{x}$
解：原式$=(\frac {x^2-4x+3}{x-3}+\frac {1}{x-3})·[\frac {(x-1)^2}{(x-1)(x-2)}-\frac {2}{x-2}]$
$=\frac {(x-2)^2}{x-3}·(\frac {x-1}{x-2}-\frac {2}{x-2})$
$=\frac {(x-2)^2}{x-3}·\frac {x-3}{x-2}$
=x-2
2. 已知 $ A = (x - 2 + \frac{3}{x + 2}) ÷ \frac{x^{2} + 2x + 1}{x + 2} $，化简 $ A $，并求当 $ x = 3 $ 时，$ A $ 的值。
答案
解：原式$=\frac{(x-2)(x+2)+3}{x+2}·\frac{x+2}{(x+1)^2}=\frac{x^2-1}{x+2}·\frac{x+2}{(x+1)^2}=\frac{(x+1)(x-1)}{x+2}·\frac{x+2}{(x+1)^2}=\frac{x-1}{x+1}$
当x=3时，$A=\frac{3-1}{3+1}=\frac{1}{2}$
当x=3时，$A=\frac{3-1}{3+1}=\frac{1}{2}$
3. (2025·苏州)先化简，再求值：$ (\frac{2}{x - 1} + 1) · \frac{x^{2} - x}{x^{2} + 2x + 1} $，其中 $ x = -2 $。
答案
解：原式$=\frac{2+x-1}{x-1}·\frac{x(x-1)}{(x+1)^2}=\frac{x+1}{x-1}·\frac{x(x-1)}{(x+1)^2}=\frac{x}{x+1}$
当x=-2时，原式$=\frac{-2}{-2+1}=2$
当x=-2时，原式$=\frac{-2}{-2+1}=2$
4. (2024·玄武区期中)先化简：$ (1 - \frac{1}{m - 2}) ÷ \frac{m^{2} - 6m + 9}{m - 2} $，然后从 $ 1 $，$ 2 $，$ 3 $ 中选一个合适的数代入求值。
答案
解：原式$=(\frac{m-2}{m-2}-\frac{1}{m-2})·\frac{m-2}{(m-3)^2}=\frac{m-3}{m-2}·\frac{m-2}{(m-3)^2}=\frac{1}{m-3}$
由题意，得$m-2≠0,m-3≠0,\therefore m≠2,m≠3$，
当m=1时，原式$=\frac{1}{1-3}=-\frac{1}{2}$
由题意，得$m-2≠0,m-3≠0,\therefore m≠2,m≠3$，
当m=1时，原式$=\frac{1}{1-3}=-\frac{1}{2}$
