《2025年名校课堂八年级数学上册人教版5四川专版》

1. 计算：
(1) $(1 - \frac{1}{x}) ÷ \frac{2x - 2}{x^2}$.
(2) $\frac{2a}{a + 1} - \frac{2a - 4}{a^2 - 1} ÷ \frac{a - 2}{a^2 - 2a + 1}$.
(3) (2024·德州) $1 - \frac{m^2 - 3m}{m^2 - 9} ÷ \frac{m + 1}{m + 3}$.
(4) $(m + 2 + \frac{1}{m}) \cdot \frac{m}{m + 1}$.
(5) $(\frac{a - b}{ab})^2 \cdot (\frac{-a}{b - a})^3 ÷ \frac{1}{a^2 - b^2}$.
(6) $\frac{2 - x}{x - 1} ÷ (x + 1 - \frac{3}{x - 1})$.

答案： 1.解:
(1)原式$=\frac {x-1}{(a-1)^{2}}\cdot \frac {x^{2}}{2a}\cdot \frac {2(a-1)}{x}=\frac {x}{a-2}=\frac {x}{a+1}-\frac {2}{a+1}=\frac {x-2}{a+1}$.
(3)原式$=1-\frac {(m+3)(m-3)}{(m+1)(m-1)}\cdot \frac {m+1}{m+3}=1-\frac {m-3}{m-1}=\frac {m-1-m+3}{m-1}=\frac {2}{m-1}$.
(4)原式$=\frac {m^{2}+2m+1}{m}\cdot \frac {m}{m+1}=\frac {(m+1)^{2}}{m}\cdot \frac {m}{m+1}=m+1$.
(5)原式$=\frac {(a-b)^{2}}{a(a+b)}\cdot \frac {a^{2}}{a^{2}+ab}\cdot \frac {(a+b)(a-b)}{(a-b)^{2}}=\frac {(a-b)^{2}}{a(a+b)}\cdot \frac {a^{2}}{a(a+b)}\cdot \frac {(a+b)(a-b)}{(a-b)^{2}}=\frac {a-b}{(a+b)^{2}}$.
(6)原式$=\frac {2-x}{x-1}÷\frac {x^{2}-4}{x-1}=-\frac {x-2}{x-1}\cdot \frac {x-1}{(x+2)(x-2)}=-\frac {1}{x+2}.$

2. (2024·湖南) 先化简，再求值：$\frac{x^2 - 4}{x^2} \cdot \frac{x}{x + 2} + \frac{3}{x}$，其中 $x = 3$.

答案： 2.解:原式$=\frac {(x+2)(x-2)}{x^{2}}\cdot \frac {x}{x+2}+\frac {3}{x}=\frac {x-2}{x}+\frac {3}{x}=\frac {x+1}{x}$.当$x=3$时,原式$=\frac {3+1}{3}=\frac {4}{3}.$

3. 先化简，再求值：$\frac{x^2}{x - 1} ÷ \frac{x}{x^2 - 1} - x$，其中 $x = -\sqrt{5}$.

答案： 3.解:原式$=\frac {x^{2}}{x-1}\cdot \frac {(x+1)(x-1)}{x}-x=x(x+1)-x=x^{2}+x-x=x^{2}$.当$x=-\sqrt {5}$时,原式$=(-\sqrt {5})^{2}=5.$