《2025年名校课堂八年级数学上册人教版云南专版》

1. 计算：
(1) (昭通昭阳区期末) $(\frac{-2m}{3n})^{2} \cdot \frac{6n}{m} ÷ (-\frac{n}{4m})$；
(2) $(1 + \frac{b}{a - b}) ÷ \frac{a}{a^{2} - b^{2}}$；
(3) $(\frac{1}{x + 1} + \frac{1}{x - 1}) \cdot (x^{2} - 1)$；
(4) $\frac{a - 2}{a + 3} ÷ \frac{a^{2} - 4}{2a + 6} - \frac{5}{a + 2}$；
(5) $(\frac{3}{x + 1} - \frac{1}{x}) ÷ \frac{2x^{2} - x}{x^{2} + 2x + 1}$；
(6) $\frac{x^{2} - y^{2}}{x} ÷ (x - \frac{2xy - y^{2}}{x})$；
(7) $(a - 1 - \frac{4a - 1}{a + 1}) ÷ \frac{a^{2} - 8a + 16}{a + 1}$。

答案： 1.
(1)解：原式$= \frac{4m^2}{9n^2} \cdot \frac{6n}{m} \cdot ( - \frac{4m}{n} )= - \frac{32m^2}{3n^2}。$
(2)解：原式$= \frac{a}{a - b} \cdot \frac{(a + b)(a - b)}{a}= a + b。$
(3)解：原式$= \frac{x - 1 + x + 1}{(x + 1)(x - 1)} \cdot (x + 1)(x - 1)= 2x。$
(4)解：原式$= \frac{a - 2}{a + 3} \cdot \frac{2(a + 3)}{(a + 2)(a - 2)}= \frac{5}{a + 2} - \frac{2}{a + 2} - \frac{5}{a + 2}= \frac{3}{a + 2}。$
(5)解：原式$= [ \frac{3x}{x(x + 1)} - \frac{x + 1}{x(x + 1)} ] \cdot \frac{x^2 + 2x + 1}{2x^2 - x}= \frac{2x - 1}{x(x + 1)} \cdot \frac{(x + 1)^2}{x(2x - 1)}= \frac{x + 1}{x^2}。$
(6)解：原式$= \frac{(x + y)(x - y)}{x} \cdot \frac{x^2 - 2xy + y^2}{x - y}= \frac{x}{(x + y)(x - y)} \cdot \frac{x}{x - y}= \frac{x}{x - y} = \frac{x + y}{x}。$
(7)解：原式$= \frac{a^2 - 1 - 4a + 1}{a + 1}= \frac{a(a - 4)}{(a - 4)^2}= \frac{a}{a - 4}。$

2. (云南中考) 先化简，再求值：$\frac{x^{2} - x}{x^{2} - 2x + 1} \cdot (x - \frac{1}{x})$，其中$x = \frac{1}{5}$。

答案： 2.解：原式$= \frac{x(x - 1)}{(x - 1)^2} - \frac{(x + 1)(x - 1)}{x}= x + 1。$当$x = \frac{1}{5}$时，原式$= \frac{6}{5}。$