登录
1. (2024·天津中考)计算$\frac{3x}{x - 1} - \frac{3}{x - 1}$的结果等于 ( )
A. 3
B. $x$
C. $\frac{x}{x - 1}$
D. $\frac{3}{x^{2} - 1}$
A. 3
B. $x$
C. $\frac{x}{x - 1}$
D. $\frac{3}{x^{2} - 1}$
答案
A
2. (2024·无锡模拟)化简$\frac{1}{a + 1} + a - 1$的结果是 ( )
A. 1
B. $\frac{a^{2}}{a^{2} - 1}$
C. $\frac{a^{2}}{a + 1}$
D. $\frac{1}{a + 1}$
A. 1
B. $\frac{a^{2}}{a^{2} - 1}$
C. $\frac{a^{2}}{a + 1}$
D. $\frac{1}{a + 1}$
答案
C
3. (2024·河北中考)已知$A$为整式，若计算$\frac{A}{xy + y^{2}} - \frac{y}{x^{2} + xy}$的结果为$\frac{x - y}{xy}$，则$A =$ ( )
A. $x$
B. $y$
C. $x + y$
D. $x - y$
A. $x$
B. $y$
C. $x + y$
D. $x - y$
答案
A
4. 化简：
(1)(2023·上海中考)$\frac{2}{1 - x} - \frac{2x}{1 - x} =$______；
(2)(2024·威海中考)$\frac{4}{x - 2} + \frac{x^{2}}{2 - x} =$______；
(3)(武汉中考)$\frac{2}{m + n} - \frac{m - 3n}{m^{2} - n^{2}} =$______.
(1)(2023·上海中考)$\frac{2}{1 - x} - \frac{2x}{1 - x} =$______；
(2)(2024·威海中考)$\frac{4}{x - 2} + \frac{x^{2}}{2 - x} =$______；
(3)(武汉中考)$\frac{2}{m + n} - \frac{m - 3n}{m^{2} - n^{2}} =$______.
答案
(1)2 (2)$-x - 2$ (3)$\frac{1}{m - n}$
5. 若$a^{2} + 5ab - b^{2} = 0$，则$\frac{b}{a} - \frac{a}{b}$的值为______.
答案
5 解析：$\because a^{2}+5ab - b^{2}=0$，$\therefore 5ab = b^{2}-a^{2}$.$\therefore$原式$=\frac{b^{2}-a^{2}}{ab}=\frac{5ab}{ab}=5$.
6. 教材P108习题T1变式 计算：
(1)(无锡中考)$\frac{a - 1}{a - b} - \frac{1 + b}{b - a}$；
(2)(连云港中考)$\frac{1}{x - 1} + \frac{x^{2} - 3x}{x^{2} - 1}$；
(3)$\frac{a^{2} - 4}{a^{2} - 4a + 4} - \frac{4a}{a^{2} - 2a}$；
(4)$\frac{4}{x + 2} - 2 + x$.
(1)(无锡中考)$\frac{a - 1}{a - b} - \frac{1 + b}{b - a}$；
(2)(连云港中考)$\frac{1}{x - 1} + \frac{x^{2} - 3x}{x^{2} - 1}$；
(3)$\frac{a^{2} - 4}{a^{2} - 4a + 4} - \frac{4a}{a^{2} - 2a}$；
(4)$\frac{4}{x + 2} - 2 + x$.
答案
(1)原式$=\frac{a - 1}{a - b}+\frac{1 + b}{a - b}=\frac{a - 1 + 1 + b}{a - b}=\frac{a + b}{a - b}$.
(2)原式$=\frac{x + 1}{x^{2}-1}+\frac{x^{2}-3x}{x^{2}-1}=\frac{x + 1 + x^{2}-3x}{x^{2}-1}=\frac{x^{2}-2x + 1}{x^{2}-1}=\frac{(x - 1)^{2}}{x^{2}-1}=\frac{(x - 1)^{2}}{(x + 1)(x - 1)}=\frac{x - 1}{x + 1}$.
(3)原式$=\frac{(a + 2)(a - 2)}{(a - 2)^{2}}-\frac{4a}{a(a - 2)}=\frac{a + 2}{a - 2}-\frac{4}{a - 2}=\frac{a + 2 - 4}{a - 2}=1$.
(4)原式$=\frac{4}{x + 2}+(x - 2)=\frac{4+(x + 2)(x - 2)}{x + 2}=\frac{x^{2}}{x + 2}$.
(2)原式$=\frac{x + 1}{x^{2}-1}+\frac{x^{2}-3x}{x^{2}-1}=\frac{x + 1 + x^{2}-3x}{x^{2}-1}=\frac{x^{2}-2x + 1}{x^{2}-1}=\frac{(x - 1)^{2}}{x^{2}-1}=\frac{(x - 1)^{2}}{(x + 1)(x - 1)}=\frac{x - 1}{x + 1}$.
(3)原式$=\frac{(a + 2)(a - 2)}{(a - 2)^{2}}-\frac{4a}{a(a - 2)}=\frac{a + 2}{a - 2}-\frac{4}{a - 2}=\frac{a + 2 - 4}{a - 2}=1$.
(4)原式$=\frac{4}{x + 2}+(x - 2)=\frac{4+(x + 2)(x - 2)}{x + 2}=\frac{x^{2}}{x + 2}$.
7. (2023·大庆中考)先化简，再求值：$\frac{2x}{x + 2} - \frac{x}{x - 2} + \frac{4x}{x^{2} - 4}$，其中$x = 1$.
答案
原式$=\frac{2x}{x + 2}-\frac{x}{x - 2}+\frac{4x}{(x + 2)(x - 2)}=\frac{2x(x - 2)}{(x + 2)(x - 2)}-\frac{x(x + 2)}{(x + 2)(x - 2)}+\frac{4x}{(x + 2)(x - 2)}=\frac{2x^{2}-4x - x^{2}-2x + 4x}{(x + 2)(x - 2)}=\frac{x^{2}-2x}{(x + 2)(x - 2)}=\frac{x(x - 2)}{(x + 2)(x - 2)}=\frac{x}{x + 2}$，当$x = 1$时，原式$=\frac{1}{1 + 2}=\frac{1}{3}$.
8. (河北中考)如图，若$x$为正整数，则表示$\frac{(x + 2)^{2}}{x^{2} + 4x + 4} - \frac{1}{x + 1}$的值的点落在 ( )
A. 段①
B. 段②
C. 段③
D. 段④
A. 段①
B. 段②
C. 段③
D. 段④
答案
B 解析：$\because\frac{(x + 2)^{2}}{x^{2}+4x + 4}-\frac{1}{x + 1}=\frac{(x + 2)^{2}}{(x + 2)^{2}}-\frac{1}{x + 1}=1-\frac{1}{x + 1}=\frac{x}{x + 1}$，又$\because x$为正整数，$\therefore\frac{1}{2}\leq\frac{x}{x + 1}＜1$. 故表示$\frac{(x + 2)^{2}}{x^{2}+4x + 4}-\frac{1}{x + 1}$的值的点落在段②. 故选 B.
9. 化简：(1)$\frac{a}{a + b - c} + \frac{b}{b - c + a} + \frac{c}{c - a - b} =$______；
(2)$\frac{1}{1 + a} + \frac{1}{1 - a} + \frac{2}{1 + a^{2}} + \frac{4}{1 + a^{4}} =$______.
(2)$\frac{1}{1 + a} + \frac{1}{1 - a} + \frac{2}{1 + a^{2}} + \frac{4}{1 + a^{4}} =$______.
答案
(1)1
(2)$\frac{8}{1 - a^{8}}$ 解析：原式$=\frac{1 - a}{(1 + a)(1 - a)}+\frac{1 + a}{(1 + a)(1 - a)}+\frac{2}{1 + a^{2}}+\frac{4}{1 + a^{4}}=\frac{2}{1 - a^{2}}+\frac{2}{1 + a^{2}}+\frac{4}{1 + a^{4}}=\frac{2(1 + a^{2})}{(1 + a^{2})(1 - a^{2})}+\frac{2(1 - a^{2})}{(1 + a^{2})(1 - a^{2})}+\frac{4}{1 + a^{4}}=\frac{4}{1 - a^{4}}+\frac{4}{1 + a^{4}}=\frac{4(1 + a^{4})}{(1 + a^{4})(1 - a^{4})}+\frac{4(1 - a^{4})}{(1 + a^{4})(1 - a^{4})}=\frac{8}{1 - a^{8}}$.
(2)$\frac{8}{1 - a^{8}}$ 解析：原式$=\frac{1 - a}{(1 + a)(1 - a)}+\frac{1 + a}{(1 + a)(1 - a)}+\frac{2}{1 + a^{2}}+\frac{4}{1 + a^{4}}=\frac{2}{1 - a^{2}}+\frac{2}{1 + a^{2}}+\frac{4}{1 + a^{4}}=\frac{2(1 + a^{2})}{(1 + a^{2})(1 - a^{2})}+\frac{2(1 - a^{2})}{(1 + a^{2})(1 - a^{2})}+\frac{4}{1 + a^{4}}=\frac{4}{1 - a^{4}}+\frac{4}{1 + a^{4}}=\frac{4(1 + a^{4})}{(1 + a^{4})(1 - a^{4})}+\frac{4(1 - a^{4})}{(1 + a^{4})(1 - a^{4})}=\frac{8}{1 - a^{8}}$.
10. (1)(2024·无锡期中)已知非零实数$x$、$y$满足$\frac{1}{y} - \frac{1}{x} = 2$，则$\frac{x - y + 4xy}{xy}$的值等于______；
(2)(2024·咸阳期中)已知$\frac{1}{x} + \frac{2}{y} = 1$，且$x + y \neq 0$，则$\frac{xy - x}{2x + 2y}$的值等于______.
(2)(2024·咸阳期中)已知$\frac{1}{x} + \frac{2}{y} = 1$，且$x + y \neq 0$，则$\frac{xy - x}{2x + 2y}$的值等于______.
答案
(1)6 解析：$\because$非零实数$x$、$y$满足$\frac{1}{y}-\frac{1}{x}=2$，$\therefore\frac{x - y + 4xy}{xy}=\frac{x}{xy}-\frac{y}{xy}+\frac{4xy}{xy}=\frac{1}{y}-\frac{1}{x}+4=2 + 4=6$.
(2)$\frac{1}{2}$ 解析：由$\frac{1}{x}+\frac{2}{y}=1$可得$\frac{y + 2x}{xy}=1$，即$xy = y + 2x$，$\therefore\frac{xy - x}{2x + 2y}=\frac{y + 2x - x}{2(x + y)}=\frac{y + x}{2(x + y)}=\frac{1}{2}$.
(2)$\frac{1}{2}$ 解析：由$\frac{1}{x}+\frac{2}{y}=1$可得$\frac{y + 2x}{xy}=1$，即$xy = y + 2x$，$\therefore\frac{xy - x}{2x + 2y}=\frac{y + 2x - x}{2(x + y)}=\frac{y + x}{2(x + y)}=\frac{1}{2}$.
