1. (2024·天津中考)计算$\frac{3x}{x - 1} - \frac{3}{x - 1}$的结果等于(
A
)
A.3
B.x
C.$\frac{x}{x - 1}$
D.$\frac{3}{x^2 - 1}$
答案：1. A 解析：$\frac{3x}{x - 1} - \frac{3}{x - 1} = \frac{3x - 3}{x - 1} = \frac{3(x - 1)}{x - 1} = 3$. 故选 A.
2. 化简$\frac{1}{a + 1} + a - 1$的结果是(
C
)
A.1
B.$\frac{a^2}{a^2 - 1}$
C.$\frac{a^2}{a + 1}$
D.$\frac{1}{a + 1}$
答案：2. C 解析：$\frac{1}{a + 1} + a - 1 = \frac{1}{a + 1} + \frac{(a + 1)(a - 1)}{a + 1} = \frac{a^{2}}{a + 1}$. 故选 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
)
A.x
B.y
C.$x + y$
D.$x - y$
答案：3. A 解析：由题意可得$\frac{A}{xy + y^{2}} = \frac{y}{x^{2} + xy} + \frac{x - y}{xy} = \frac{y^{2} + (x - y)(x + y)}{xy(x + y)} =\frac{x^{2}}{xy(x + y)} = \frac{x}{y(x + y)}$，$\because \frac{A}{xy + y^{2}} = \frac{x}{y(x + y)}$，$\therefore A = x$. 故选 A.
4. 化简：
(1) (2025·新疆中考改编)$\frac{x}{x - 2y} - \frac{2y}{x - 2y} =$
1
。
(2) (2025·河南中考改编)$\frac{x^2 - 2}{x - 1} - \frac{1}{1 - x} =$
$x + 1$
。
(3) (武汉中考)$\frac{2}{m + n} - \frac{m - 3n}{m^2 - n^2} =$
$\frac{1}{m - n}$
。
答案：4. (1)1 解析：原式$= \frac{x - 2y}{x - 2y} = 1$.
(2)$x + 1$ 解析：原式$= \frac{x^{2} - 2}{x - 1} + \frac{1}{x - 1} = \frac{(x + 1)(x - 1)}{x - 1} = x + 1$.
(3)$\frac{1}{m - n}$ 解析：原式$= \frac{2(m - n)}{m^{2} - n^{2}} - \frac{m - 3n}{m^{2} - n^{2}} = \frac{m + n}{m^{2} - n^{2}} = \frac{1}{m - n}$.
5. 若$a^2 + 5ab - b^2 = 0$，则$\frac{b}{a} - \frac{a}{b}$的值为
5
。
答案：5. 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. 教材变式 计算：
(1)$\frac{a - 1}{a - b} - \frac{1 + b}{b - a}$；
(2)$\frac{1}{x - 1} + \frac{x^2 - 3x}{x^2 - 1}$；
(3)$\frac{a^2 - 4}{a^2 - 4a + 4} - \frac{4a}{a^2 - 2a}$；
(4)$\frac{4}{x + 2} - 2 + x$。
答案：6. (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$= \frac{4}{x + 2} + (x - 2) = \frac{4 + (x + 2)(x - 2)}{x + 2} = \frac{x^{2}}{x + 2}$.
7. (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：7. 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. (2025·周口模拟)若分式$\frac{a - b}{a + b} + \frac{a + b}{a - b}$有意义，则下列关于运算结果的说法中正确的是(
C
)
A.不可能是负数
B.不可能是正数
C.不可能是0
D.有可能是0
答案：8. C 解析：$\frac{a - b}{a + b} + \frac{a + b}{a - b} = \frac{(a - b)^{2}}{(a + b)(a - b)} + \frac{(a + b)^{2}}{(a + b)(a - b)} =\frac{a^{2} + b^{2} - 2ab + a^{2} + b^{2} + 2ab}{(a + b)(a - b)} = \frac{2(a^{2} + b^{2})}{(a + b)(a - b)}$，$\because$ 分式$\frac{a - b}{a + b} + \frac{a + b}{a - b}$有意义，$\therefore \{\begin{array}{l}a + b ≠ 0,\\a - b ≠ 0,\end{array} $ $\therefore a ≠ \pm b$，$\therefore$ 不存在$a = b = 0$这种情形，$\therefore 2(a^{2} + b^{2}) ＞ 0$. 又$\because (a + b)(a - b)$的值可以为负，也可以为正，$\therefore \frac{2(a^{2} + b^{2})}{(a + b)(a - b)}$的值可以为负，也可以为正，但不可以为 0，故选 C.
9. 化简：(1)$\frac{a}{a + b - c} + \frac{b}{b - c + a} + \frac{c}{c - a - b} =$
1
；
(2)$\frac{1}{1 + a} + \frac{1}{1 - a} + \frac{2}{1 + a^2} + \frac{4}{1 + a^4} =$
$\frac{8}{1 - a^{8}}$
。
答案：9. (1)1 解析：原式$= \frac{a}{a + b - c} + \frac{b}{a + b - c} - \frac{c}{a + b - c} = \frac{a + b - c}{a + b - c} = 1$.
(2)$\frac{8}{1 - a^{8}}$ 解析：原式$= \frac{1 - a}{(1 + a)(1 - a)} + \frac{1 + a}{(1 + a)(1 - a)} + \frac{2}{1 + a^{2}} +\frac{4}{1 + a^{4}} = \frac{2}{1 - a^{2}} + \frac{2}{1 + a^{2}} + \frac{4}{1 + a^{4}} = \frac{2(1 + a^{2})}{(1 + a^{2})(1 - a^{2})} + \frac{2(1 - a^{2})}{(1 + a^{2})(1 - a^{2})} +\frac{4}{1 + a^{4}} = \frac{4}{1 - a^{4}} + \frac{4}{1 + a^{4}} = \frac{4(1 + a^{4})}{(1 + a^{4})(1 - a^{4})} + \frac{4(1 - a^{4})}{(1 + a^{4})(1 - a^{4})} = \frac{8}{1 - a^{8}}$.
10. (1)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
6
；
答案：10. (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) (2025·泰州期中)已知$\frac{1}{x} + \frac{2}{y} = 1$，且$x + y ≠ 0$，则$\frac{xy - x}{2x + 2y}$的值等于
$\frac{1}{2}$
。
答案：(2)$\frac{1}{2}$ 解析：由$\frac{1}{x} + \frac{2}{y} = 1$可得$\frac{y + 2x}{xy} = 1$，即$xy = y + 2x$，$\therefore \frac{xy - x}{2x + 2y} = \frac{y + 2x - x}{2(x + y)} = \frac{y + x}{2(x + y)} = \frac{1}{2}$.
