2026年初中运算计算升级卡八年级上册人教版第98页答案
2. 先化简,再求值:$\frac{x^2 -1}{x^2 +x} ÷ \frac{x^2 -2x +1}{x}$,其中$x=3.$

答案

原式=$\frac{1}{x-1}$. 当x=3时,原式=$\frac{1}{2}$.
1. 计算:
(1) $\frac{m+2}{m+1} - \frac{1}{m+1}$;
(2) $\frac{2a}{a-2} + \frac{4}{2-a}$;
(3) $\frac{a}{a^2 -25b^2} - \frac{1}{2a -10b}$;
(4) $\frac{4x}{x^2 - y^2} - \frac{2}{y + x}$;
(5) $\frac{x^2 +2x}{x^2 -4} - \frac{2}{x -2}$;
(6) $( \frac{1}{x} - \frac{1}{y} ) · \frac{xy}{x^2 - y^2}$;
(7) $( \frac{3a}{a-3} - \frac{a}{a+3} ) · \frac{a^2 -9}{a}$;
(8) $\frac{a-1}{a} ÷ ( a - \frac{1}{a} )$;
(9) $( \frac{x+2}{x^2 -2x} - \frac{x-1}{x^2 -4x +4} ) ÷ \frac{4 -x}{x}$;
(10) $\frac{2x}{x+1} - \frac{2x+6}{x^2 -1} ÷ \frac{x+3}{x^2 -2x +1}$;

答案

(1) 1 (2) 2 (3) $\frac{1}{2a+10b}$ (4) $\frac{2}{x-y}$ (5) 1 (6) $-\frac{1}{x+y}$ (7) $2a+12$ (8) $\frac{1}{a+1}$ (9) $-\frac{1}{x^2-4x+4}$ (10) $\frac{2}{x+1}$