登录
1. 计算:
(1)$\frac{2a}{a+1}-\frac{2a-4}{a^2-1}÷\frac{a-2}{a^2-2a+1}$；
(2)$( \frac{a-b}{ab} )^2 · ( \frac{-a}{b-a} )^3 ÷ \frac{1}{a^2 - b^2}$；
(3)$( \frac{2a-1}{a^2 - a} - \frac{a}{a - 1} ) ÷ \frac{a^2 - 1}{a}$；
(4)$\frac{4}{2 - a} - a - 2 + \frac{a^2}{a - 2}$；
(5)$\frac{2 - x}{x - 1} ÷ ( x + 1 - \frac{3}{x - 1} )$；
(6)$\frac{2a}{a + 1} - \frac{2a - 4}{a^2 - 1} ÷ \frac{a - 2}{a^2 - 2a + 1}$；
(7)$( 1 + \frac{2}{x - 1} ) ÷ \frac{x + 1}{x^2 - 2x + 1}$。
(1)$\frac{2a}{a+1}-\frac{2a-4}{a^2-1}÷\frac{a-2}{a^2-2a+1}$；
(2)$( \frac{a-b}{ab} )^2 · ( \frac{-a}{b-a} )^3 ÷ \frac{1}{a^2 - b^2}$；
(3)$( \frac{2a-1}{a^2 - a} - \frac{a}{a - 1} ) ÷ \frac{a^2 - 1}{a}$；
(4)$\frac{4}{2 - a} - a - 2 + \frac{a^2}{a - 2}$；
(5)$\frac{2 - x}{x - 1} ÷ ( x + 1 - \frac{3}{x - 1} )$；
(6)$\frac{2a}{a + 1} - \frac{2a - 4}{a^2 - 1} ÷ \frac{a - 2}{a^2 - 2a + 1}$；
(7)$( 1 + \frac{2}{x - 1} ) ÷ \frac{x + 1}{x^2 - 2x + 1}$。
答案
(1)解:原式=$\frac{2a}{a+1}-\frac{2(a-2)}{(a-1)(a+1)}·\frac{(a-1)^2}{a-2}=\frac{2a}{a+1}-\frac{2(a-1)}{a+1}=\frac{2}{a+1}$.
(2)解:原式=$\frac{(a-b)^2}{a^2b^2}·\frac{a^3}{(a-b)^3}·(a+b)(a-b)=\frac{a(a+b)}{b^2}=\frac{a^2+ab}{b^2}$.
(3)解:原式=$[\frac{2a-1}{a(a-1)}-\frac{a^2}{a(a-1)}]÷\frac{(a+1)(a-1)}{a}=\frac{2a-1-a^2}{a(a-1)}·\frac{a}{(a+1)(a-1)}=\frac{-(a-1)^2}{a(a-1)}·\frac{a}{(a+1)(a-1)}=-\frac{1}{a+1}$.
(4)解:原式=$-\frac{4}{a-2}-(a+2)+\frac{a^2}{a-2}=-\frac{4}{a-2}-\frac{a^2-4}{a-2}+\frac{a^2}{a-2}=\frac{-4-a^2+4+a^2}{a-2}=0$.
(5)解:原式=$\frac{2-x}{x-1}÷\frac{x^2-4}{x-1}=-\frac{x-2}{x-1}·\frac{x-1}{(x+2)(x-2)}=-\frac{1}{x+2}$.
(6)解:原式=$\frac{2a}{a+1}-\frac{2(a-2)}{(a-1)(a+1)}·\frac{(a-1)^2}{a-2}=\frac{2a}{a+1}-\frac{2(a-1)}{a+1}=\frac{2}{a+1}$.
(7)解:原式=$\frac{x-1+2}{x-1}·\frac{(x-1)^2}{x+1}=\frac{x+1}{x-1}·\frac{(x-1)^2}{x+1}=x-1$.
(2)解:原式=$\frac{(a-b)^2}{a^2b^2}·\frac{a^3}{(a-b)^3}·(a+b)(a-b)=\frac{a(a+b)}{b^2}=\frac{a^2+ab}{b^2}$.
(3)解:原式=$[\frac{2a-1}{a(a-1)}-\frac{a^2}{a(a-1)}]÷\frac{(a+1)(a-1)}{a}=\frac{2a-1-a^2}{a(a-1)}·\frac{a}{(a+1)(a-1)}=\frac{-(a-1)^2}{a(a-1)}·\frac{a}{(a+1)(a-1)}=-\frac{1}{a+1}$.
(4)解:原式=$-\frac{4}{a-2}-(a+2)+\frac{a^2}{a-2}=-\frac{4}{a-2}-\frac{a^2-4}{a-2}+\frac{a^2}{a-2}=\frac{-4-a^2+4+a^2}{a-2}=0$.
(5)解:原式=$\frac{2-x}{x-1}÷\frac{x^2-4}{x-1}=-\frac{x-2}{x-1}·\frac{x-1}{(x+2)(x-2)}=-\frac{1}{x+2}$.
(6)解:原式=$\frac{2a}{a+1}-\frac{2(a-2)}{(a-1)(a+1)}·\frac{(a-1)^2}{a-2}=\frac{2a}{a+1}-\frac{2(a-1)}{a+1}=\frac{2}{a+1}$.
(7)解:原式=$\frac{x-1+2}{x-1}·\frac{(x-1)^2}{x+1}=\frac{x+1}{x-1}·\frac{(x-1)^2}{x+1}=x-1$.
2. 先化简，再求值：$( \dfrac{3}{x-1} - x - 1 ) ÷ \dfrac{x^2 - 4x + 4}{x - 1}$，其中 $x=3$。
答案
解:原式=$\frac{3-(x^2-1)}{x-1}·\frac{x-1}{(x-2)^2}=-\frac{(x-2)(x+2)}{x-1}·\frac{x-1}{(x-2)^2}=-\frac{x+2}{x-2}$.当x=3时,原式=$-\frac{x+2}{x-2}=-\frac{3+2}{3-2}=-5$.
