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1. 计算：
(1)$2 - \frac{3x + y}{x - 2y} \div \frac{9x^{2} + 6xy + y^{2}}{x^{2} - 4y^{2}}$；
(2)(2023·苏州期末)$\frac{x + 2}{x^{2} - 9} \div (1 - \frac{1}{x + 3})$.
(3)$(x - \frac{y^{2}}{x}) \cdot \frac{y}{x + y} - y$；
(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)$2 - \frac{3x + y}{x - 2y} \div \frac{9x^{2} + 6xy + y^{2}}{x^{2} - 4y^{2}}$；
(2)(2023·苏州期末)$\frac{x + 2}{x^{2} - 9} \div (1 - \frac{1}{x + 3})$.
(3)$(x - \frac{y^{2}}{x}) \cdot \frac{y}{x + y} - y$；
(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：(1) 原式$=2 - \frac{3x + y}{x - 2y} \cdot \frac{(x + 2y)(x - 2y)}{(3x + y)^2} = 2 - \frac{x + 2y}{3x + y} = \frac{2(3x + y) - (x + 2y)}{3x + y} = \frac{6x + 2y - x - 2y}{3x + y} = \frac{5x}{3x + y}$。
(2) 原式$=\frac{x + 2}{(x + 3)(x - 3)} \div \frac{x + 3 - 1}{x + 3} = \frac{x + 2}{(x + 3)(x - 3)} \div \frac{x + 2}{x + 3} = \frac{x + 2}{(x + 3)(x - 3)} \cdot \frac{x + 3}{x + 2} = \frac{1}{x - 3}$。
(3) 原式$=\frac{(x + y)(x - y)}{x} \cdot \frac{y}{x + y} - y = \frac{y(x - y)}{x} - \frac{xy}{x} = -\frac{y^2}{x}$。
(4) 原式$=(\frac{x^2 - 4x + 3}{x - 3} + \frac{1}{x - 3}) \cdot [\frac{(x - 1)^2}{(x - 1)(x - 2)} - \frac{2}{x - 2}] = \frac{(x - 2)^2}{x - 3} \cdot (\frac{x - 1}{x - 2} - \frac{2}{x - 2}) = \frac{(x - 2)^2}{x - 3} \cdot \frac{x - 3}{x - 2} = x - 2$。
(2) 原式$=\frac{x + 2}{(x + 3)(x - 3)} \div \frac{x + 3 - 1}{x + 3} = \frac{x + 2}{(x + 3)(x - 3)} \div \frac{x + 2}{x + 3} = \frac{x + 2}{(x + 3)(x - 3)} \cdot \frac{x + 3}{x + 2} = \frac{1}{x - 3}$。
(3) 原式$=\frac{(x + y)(x - y)}{x} \cdot \frac{y}{x + y} - y = \frac{y(x - y)}{x} - \frac{xy}{x} = -\frac{y^2}{x}$。
(4) 原式$=(\frac{x^2 - 4x + 3}{x - 3} + \frac{1}{x - 3}) \cdot [\frac{(x - 1)^2}{(x - 1)(x - 2)} - \frac{2}{x - 2}] = \frac{(x - 2)^2}{x - 3} \cdot (\frac{x - 1}{x - 2} - \frac{2}{x - 2}) = \frac{(x - 2)^2}{x - 3} \cdot \frac{x - 3}{x - 2} = x - 2$。
2. 已知$A = (x - 2 + \frac{3}{x + 2}) \div \frac{x^{2} + 2x + 1}{x + 2}$. 化简$A$；并求当$x = 3$时，$A$的值.
答案
解：原式$=\frac{(x - 2)(x + 2) + 3}{x + 2} \cdot \frac{x + 2}{(x + 1)^2} = \frac{x^2 - 1}{x + 2} \cdot \frac{x + 2}{(x + 1)^2} = \frac{(x + 1)(x - 1)}{x + 2} \cdot \frac{x + 2}{(x + 1)^2} = \frac{x - 1}{x + 1}$，
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. (2023·高港区期末)先化简$(\frac{3}{a + 1} - a + 1) \div \frac{a^{2} - 4a + 4}{a + 1}$，然后从$-2 \leq a \leq 2$的范围内选择一个合适的整数作为$a$的值代入求值.
答案
解：原式$=[\frac{3}{a + 1} - \frac{(a - 1)(a + 1)}{a + 1}] \cdot \frac{a + 1}{(a - 2)^2} = \frac{3 - a^2 + 1}{a + 1} \cdot \frac{a + 1}{(a - 2)^2} = \frac{4 - a^2}{(a - 2)^2} = \frac{(2 - a)(2 + a)}{(2 - a)^2} = \frac{a + 2}{2 - a}$，
由分式有意义的条件可知$a \neq -1,a \neq 2$，
$\therefore$故$a$可取$a = 0$，
$\therefore$原式$=\frac{2}{2} = 1$。
由分式有意义的条件可知$a \neq -1,a \neq 2$，
$\therefore$故$a$可取$a = 0$，
$\therefore$原式$=\frac{2}{2} = 1$。
4. (2024·泰州月考)先化简，再求值：$(\frac{x^{2}}{x - 1} - x + 1) \div \frac{4x^{2} - 4x + 1}{1 - x}$，其中$x$满足$x^{2} + 2x - 3 = 0$.
答案
解：原式$=\frac{x^2 - (x - 1)^2}{x - 1} \cdot \frac{1 - x}{(2x - 1)^2} = \frac{2x - 1}{x - 1} \cdot \frac{1 - x}{(2x - 1)^2} = -\frac{1}{2x - 1}$，
由$x^2 + 2x - 3 = 0$，解得$x_1 = -3,x_2 = 1$，
$\because x \neq 1$，
$\therefore$当$x = -3$时，原式$=-\frac{1}{2\times(-3) - 1} = \frac{1}{7}$。
由$x^2 + 2x - 3 = 0$，解得$x_1 = -3,x_2 = 1$，
$\because x \neq 1$，
$\therefore$当$x = -3$时，原式$=-\frac{1}{2\times(-3) - 1} = \frac{1}{7}$。
