《2025年名校课堂八年级数学上册青岛版》

1. 解下列分式方程：
(1)(遵义中考)$\frac {1}{x - 2} = \frac {3}{2x - 3}$；
(2)(镇江中考)$\frac {2}{x - 2} = \frac {1 + x}{x - 2} + 1$；
(3)(泰安肥城市期末)$\frac {7}{x - 2} + \frac {x}{2 - x} = 4$；
(4)$\frac {4}{x^{2} - 1} + 2 = \frac {2x}{x - 1}$；
(5)$\frac {14}{x + 8} = \frac {4}{x} + \frac {10}{3x + 24}$；
(6)$\frac {x}{x - 1} - 1 = \frac {3}{(x - 1)(x + 2)}$.

答案： 1.解：
(1)方程两边乘$(x - 2)(2x - 3)$，得$2x - 3 = 3(x - 2)$.解得$x = 3$.检验：当$x = 3$时，$(x - 2)(2x - 3)\neq 0$.所以原分式方程的解为$x = 3$.
(2)方程两边乘$(x - 2)$，得$2 = 1 + x + x - 2$.解得$x = \frac{3}{2}$.检验：当$x = \frac{3}{2}$时，$x - 2\neq 0$.所以原分式方程的解为$x = \frac{3}{2}$.
(3)方程两边乘$(x - 2)$，得$7 - x = 4(x - 2)$.解得$x = 3$.检验：当$x = 3$时，$x - 2\neq 0$.所以原分式方程的解为$x = 3$.
(4)去分母，得$4 + 2x^{2} - 2 = 2x(x + 1)$，解得$x = 1$.检验：当$x = 1$时，$x^{2} - 1 = 0$.所以$x = 1$是原方程的增根.所以原方程无解.
(5)方程两边乘$3x(x + 8)$，得$42x = 12(x + 8) + 10x$.解得$x = \frac{24}{5}$.检验：当$x = \frac{24}{5}$时，$3x(x + 8)\neq 0$.
所以原分式方程的解为$x = \frac{24}{5}$.
(6)方程两边乘$(x - 1)(x + 2)$，得$x(x + 2) - (x - 1)(x + 2) = 3$.解得$x = 1$.检验：当$x = 1$时，$(x - 1)(x + 2) = 0$.所以$x = 1$是原方程的增根.所以原方程无解.

2. 解分式方程：$\frac {1}{x - 4} - \frac {1}{x - 5} = \frac {1}{x - 7} - \frac {1}{x - 8}$.

答案： 2.解：$\frac{(x - 5) - (x - 4)}{(x - 4)(x - 5)} = \frac{(x - 8) - (x - 7)}{(x - 7)(x - 8)}$，$\frac{-1}{x^{2} - 9x + 20} = \frac{-1}{x^{2} - 15x + 56}$，$6x = 36$，$x = 6$.经检验，$x = 6$是原分式方程的解.

3. 已知关于$x$的分式方程$\frac {2}{x + 4} = \frac {m}{x}$与分式方程$\frac {3}{2x} = \frac {1}{x - 1}$的解相同，求$m^{2} - 2m$的值.

答案： 3.解：解分式方程$\frac{3}{2x} = \frac{1}{x - 1}$，得$x = 3$.将$x = 3$代入$\frac{2}{x + 4} = \frac{m}{x}$，得$\frac{2}{7} = \frac{m}{3}$，解得$m = \frac{6}{7}$.所以$m^{2} - 2m = (\frac{6}{7})^{2} - 2×\frac{6}{7} = -\frac{48}{49}$.