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7. 计算.
(1) $\frac{x + 2}{x^{2}-2x}-\frac{x - 1}{x^{2}-4x + 4}$；
(2) $\frac{x^{2}}{x - 1}-x - 1$；
(3) $(xy - x^{2})(\frac{1}{x}+\frac{1}{y - x})$；
(4) $(\frac{x^{2}-4x + 4}{x^{2}-4}-\frac{x}{x + 2})÷\frac{x - 1}{x + 2}$.
(1) $\frac{x + 2}{x^{2}-2x}-\frac{x - 1}{x^{2}-4x + 4}$；
(2) $\frac{x^{2}}{x - 1}-x - 1$；
(3) $(xy - x^{2})(\frac{1}{x}+\frac{1}{y - x})$；
(4) $(\frac{x^{2}-4x + 4}{x^{2}-4}-\frac{x}{x + 2})÷\frac{x - 1}{x + 2}$.
答案
7.(1)解:原式=$\frac{x + 2}{x(x - 2)} - \frac{x - 1}{(x - 2)²}$
=$\frac{x² - 4 - x² + x}{x(x - 2)²}$
=$\frac{x - 4}{x(x - 2)}$
(2)解:原式=$\frac{x² - (x² - 1)}{x - 1}$
=$\frac{1}{x - 1}$
(3)解:原式=x(y - x)·$\frac{y - x + x}{x(y - x)}$
=y
(4)解:原式=($\frac{x - 2}{x + 2} - \frac{x}{x + 2}$)·$\frac{x + 2}{x - 1}$
=$\frac{x - 2 - x}{x - 1}$
=-$\frac{2}{x - 1}$
=$\frac{x² - 4 - x² + x}{x(x - 2)²}$
=$\frac{x - 4}{x(x - 2)}$
(2)解:原式=$\frac{x² - (x² - 1)}{x - 1}$
=$\frac{1}{x - 1}$
(3)解:原式=x(y - x)·$\frac{y - x + x}{x(y - x)}$
=y
(4)解:原式=($\frac{x - 2}{x + 2} - \frac{x}{x + 2}$)·$\frac{x + 2}{x - 1}$
=$\frac{x - 2 - x}{x - 1}$
=-$\frac{2}{x - 1}$
若$ab = 1$，求$\frac{a}{a + 1}+\frac{b}{b + 1}$的值.
答案
解:原式=$\frac{ab + a + ab + b}{(a + 1)(b + 1)}$
=$\frac{2ab + a + b}{ab + a + b + 1}$
当ab = 1时
原式=$\frac{a + b + 2}{a + b + 2}$
=1
=$\frac{2ab + a + b}{ab + a + b + 1}$
当ab = 1时
原式=$\frac{a + b + 2}{a + b + 2}$
=1
