《2025年名校课堂八年级数学上册人教版安徽专版》

1. 计算：
(1) $(1 - \frac{1}{x}) ÷ \frac{2x - 2}{x^2}$.
(2) $(\frac{2a - 1}{a^2 - a} - \frac{a}{a - 1}) ÷ \frac{a^2 - 1}{a}$.
(3) $(\frac{a - b}{ab})^2 \cdot (\frac{-a}{b - a})^3 ÷ \frac{1}{a^2 - b^2}$.
(4) $\frac{2 - x}{x - 1} ÷ (x + 1 - \frac{3}{x - 1})$.

答案： 1. 解:
(1)原式$=\frac {x-1}{x}\cdot \frac {x^{2}}{2(x-1)}=\frac {x}{2}.$
(2)原式$=[\frac {2a-1}{(a+1)(a-1)}-\frac {a^{2}}{(a+1)(a-1)}]÷\frac {a}{a-1}=\frac {2a-1-a^{2}}{(a+1)(a-1)}\cdot \frac {a-1}{a}=\frac {-(a-1)^{2}}{(a+1)(a-1)}\cdot \frac {a-1}{a}=-\frac {1}{a+1}.$
(3)原式$=\frac {(a-b)^{2}}{a^{2}b^{2}}\cdot \frac {a^{3}}{(a-b)^{3}}\cdot (a+b)(a-b)=\frac {a(a+b)}{b^{2}}=\frac {a^{2}+ab}{b^{2}}.$
(4)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. 先化简，再求值：$\frac{x^2}{x - 1} ÷ \frac{x}{x^2 - 1} - x$，其中 $x = -\sqrt{5}$.

答案： 2. 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$=(-\sqrt {5})^{2}=5.$

3. (2024·淮南潘集区期末改编)先化简，再求值：$\frac{x - 2}{x + 3} \cdot \frac{2x + 6}{x^2 - 4} - \frac{5}{x + 2}$，请选一个合适的 $x$ 值代入求值.

答案： 3. 解:原式$=\frac {x-2}{x+3}\cdot \frac {2(x+3)}{(x+2)(x-2)}-\frac {5}{x+2}=\frac {2}{x+2}-\frac {5}{x+2}=\frac {2-5}{x+2}=-\frac {3}{x+2}.\because x+3≠0$且$x+2≠0$且$x-2≠0,\therefore x$可以取1,当$x=1$时,原式$=-\frac {3}{1+2}=-1$.(答案不唯一)

4. (2024·广元)先化简，再求值：$\frac{a}{a - b} ÷ \frac{a^2 - b^2}{a^2 - 2ab + b^2} - \frac{a - b}{a + b}$，其中 $a$，$b$ 满足 $b - 2a = 0$.

答案： 4. 解:原式$=\frac {a-b}{a-b}\cdot \frac {(a-b)^{2}}{(a+b)(a-b)}-\frac {a-b}{a+b}=\frac {a-b}{a+b}-\frac {a-b}{a+b}=\frac {b}{a+b}.\because b-2a=0,\therefore b=2a.\therefore $原式$=\frac {2a}{a+2a}=\frac {2}{3}.$

5. (2024·淮南期末)先化简，再求值：$(\frac{x + 2}{x^2 - 2x} - \frac{x - 1}{x^2 - 4x + 4}) ÷ \frac{x - 4}{x^3 - 2x^2}$，其中 $x$ 与 $2$，$3$ 构成 $\triangle ABC$ 的三边，且 $x$ 为整数.

答案： 5. 解:原式$=[\frac {x^{2}-4}{x(x-2)^{2}}-\frac {x^{2}-x}{x(x-2)^{2}}]\cdot \frac {x^{2}(x-2)}{x-4}=\frac {x-4}{x(x-2)^{2}}\cdot \frac {x^{2}(x-2)}{x-4}=\frac {x}{x-2}.\because x$与2,3构成$\triangle ABC$的三边,且x为整数,$\therefore 3-2＜x＜3+2$,即$1＜x＜5.\therefore x=2$或$x=3$或$x=4$.当$x=2$或$x=4$时,原式没有意义,故$x=3$.当$x=3$时,原式$=\frac {3}{3-2}=3.$