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例1 计算：
(1)$(\frac {x + 2}{x^{2}-2x}-\frac {x - 1}{x^{2}-4x + 4})÷\frac {4 - x}{x}$；
(2)$\frac {x}{x - y}·\frac {y^{2}}{x + y}-\frac {x^{4}y}{x^{4}-y^{4}}÷\frac {x^{2}}{x^{2}+y^{2}}$。
(1)$(\frac {x + 2}{x^{2}-2x}-\frac {x - 1}{x^{2}-4x + 4})÷\frac {4 - x}{x}$；
(2)$\frac {x}{x - y}·\frac {y^{2}}{x + y}-\frac {x^{4}y}{x^{4}-y^{4}}÷\frac {x^{2}}{x^{2}+y^{2}}$。
答案
(1) $-\dfrac{1}{(x - 2)^2}$；(2) $-\dfrac{xy}{x + y}$
解析
(1)
$\begin{aligned}&\left(\frac{x + 2}{x^2 - 2x} - \frac{x - 1}{x^2 - 4x + 4}\right) ÷ \frac{4 - x}{x}\\=&\left[\frac{x + 2}{x(x - 2)} - \frac{x - 1}{(x - 2)^2}\right] · \frac{x}{4 - x}\\=&\left[\frac{(x + 2)(x - 2) - x(x - 1)}{x(x - 2)^2}\right] · \frac{x}{4 - x}\\=&\left[\frac{x^2 - 4 - x^2 + x}{x(x - 2)^2}\right] · \frac{x}{4 - x}\\=&\frac{x - 4}{x(x - 2)^2} · \frac{x}{4 - x}\\=&\frac{-(4 - x)}{x(x - 2)^2} · \frac{x}{4 - x}\\=&-\frac{1}{(x - 2)^2}\end{aligned}$
(2)
$\begin{aligned}&\frac{x}{x - y} · \frac{y^2}{x + y} - \frac{x^4y}{x^4 - y^4} ÷ \frac{x^2}{x^2 + y^2}\\=&\frac{xy^2}{(x - y)(x + y)} - \frac{x^4y}{(x^2 + y^2)(x^2 - y^2)} · \frac{x^2 + y^2}{x^2}\\=&\frac{xy^2}{x^2 - y^2} - \frac{x^4y}{(x^2 - y^2)x^2}\\=&\frac{xy^2}{x^2 - y^2} - \frac{x^2y}{x^2 - y^2}\\=&\frac{xy^2 - x^2y}{x^2 - y^2}\\=&\frac{xy(y - x)}{(x - y)(x + y)}\\=&-\frac{xy}{x + y}\end{aligned}$
$\begin{aligned}&\left(\frac{x + 2}{x^2 - 2x} - \frac{x - 1}{x^2 - 4x + 4}\right) ÷ \frac{4 - x}{x}\\=&\left[\frac{x + 2}{x(x - 2)} - \frac{x - 1}{(x - 2)^2}\right] · \frac{x}{4 - x}\\=&\left[\frac{(x + 2)(x - 2) - x(x - 1)}{x(x - 2)^2}\right] · \frac{x}{4 - x}\\=&\left[\frac{x^2 - 4 - x^2 + x}{x(x - 2)^2}\right] · \frac{x}{4 - x}\\=&\frac{x - 4}{x(x - 2)^2} · \frac{x}{4 - x}\\=&\frac{-(4 - x)}{x(x - 2)^2} · \frac{x}{4 - x}\\=&-\frac{1}{(x - 2)^2}\end{aligned}$
(2)
$\begin{aligned}&\frac{x}{x - y} · \frac{y^2}{x + y} - \frac{x^4y}{x^4 - y^4} ÷ \frac{x^2}{x^2 + y^2}\\=&\frac{xy^2}{(x - y)(x + y)} - \frac{x^4y}{(x^2 + y^2)(x^2 - y^2)} · \frac{x^2 + y^2}{x^2}\\=&\frac{xy^2}{x^2 - y^2} - \frac{x^4y}{(x^2 - y^2)x^2}\\=&\frac{xy^2}{x^2 - y^2} - \frac{x^2y}{x^2 - y^2}\\=&\frac{xy^2 - x^2y}{x^2 - y^2}\\=&\frac{xy(y - x)}{(x - y)(x + y)}\\=&-\frac{xy}{x + y}\end{aligned}$
例2 先化简,后求值：
$(1+\frac {1}{x - 2})÷\frac {x^{2}-2x + 1}{x^{2}-4}$,其中$x = - 5$。
$(1+\frac {1}{x - 2})÷\frac {x^{2}-2x + 1}{x^{2}-4}$,其中$x = - 5$。
答案
$(1+\frac{1}{x - 2})÷\frac{x^{2}-2x + 1}{x^{2}-4}$
$=(\frac{x - 2}{x - 2}+\frac{1}{x - 2})÷\frac{(x - 1)^{2}}{(x + 2)(x - 2)}$
$=\frac{x - 1}{x - 2}·\frac{(x + 2)(x - 2)}{(x - 1)^{2}}$
$=\frac{x + 2}{x - 1}$
当$x = -5$时，原式$=\frac{-5 + 2}{-5 - 1}=\frac{-3}{-6}=\frac{1}{2}$
$=(\frac{x - 2}{x - 2}+\frac{1}{x - 2})÷\frac{(x - 1)^{2}}{(x + 2)(x - 2)}$
$=\frac{x - 1}{x - 2}·\frac{(x + 2)(x - 2)}{(x - 1)^{2}}$
$=\frac{x + 2}{x - 1}$
当$x = -5$时，原式$=\frac{-5 + 2}{-5 - 1}=\frac{-3}{-6}=\frac{1}{2}$
化简：$(\frac {x}{x + 1}+\frac {x}{x - 1})\cdot\frac {x^{2}-1}{x}$。下面是甲、乙两同学的部分运算过程：
甲同学：解：原式$=[\frac {x(x - 1)}{(x + 1)(x - 1)}+\frac {x(x + 1)}{(x - 1)(x + 1)}]\cdot\frac {x^{2}-1}{x}$
乙同学：解：原式$=\frac {x}{x + 1}\cdot\frac {x^{2}-1}{x}+\frac {x}{x - 1}\cdot\frac {x^{2}-1}{x}$
(1)甲同学解法的依据是
①等式的基本性质；②分式的基本性质；③乘法分配律；④乘法交换律。
(2)请选择一种解法,写出完整的解答过程。
甲同学：解：原式$=[\frac {x(x - 1)}{(x + 1)(x - 1)}+\frac {x(x + 1)}{(x - 1)(x + 1)}]\cdot\frac {x^{2}-1}{x}$
乙同学：解：原式$=\frac {x}{x + 1}\cdot\frac {x^{2}-1}{x}+\frac {x}{x - 1}\cdot\frac {x^{2}-1}{x}$
(1)甲同学解法的依据是
②
,乙同学解法的依据是____③
。(填序号)①等式的基本性质；②分式的基本性质；③乘法分配律；④乘法交换律。
(2)请选择一种解法,写出完整的解答过程。
答案
(1)②；③
(2)选择乙同学的解法：
原式$=\frac{x}{x+1}\cdot\frac{x^{2}-1}{x}+\frac{x}{x-1}\cdot\frac{x^{2}-1}{x}$
$=\frac{x}{x+1}\cdot\frac{(x+1)(x-1)}{x}+\frac{x}{x-1}\cdot\frac{(x+1)(x-1)}{x}$
$=(x - 1)+(x + 1)$
$=x - 1 + x + 1$
$=2x$
(2)选择乙同学的解法：
原式$=\frac{x}{x+1}\cdot\frac{x^{2}-1}{x}+\frac{x}{x-1}\cdot\frac{x^{2}-1}{x}$
$=\frac{x}{x+1}\cdot\frac{(x+1)(x-1)}{x}+\frac{x}{x-1}\cdot\frac{(x+1)(x-1)}{x}$
$=(x - 1)+(x + 1)$
$=x - 1 + x + 1$
$=2x$
