《2025年期末考向标八年级数学上册冀教版》

14. (邢台月考)由下表数据可知，$a + b =$0.

答案： 14．$0$ [解析]观察表可知$\frac{x - 1}{x} = 2$，方程两边同乘$x$，得$x - 1 = 2x$，解得$x = -1$．经检验，$x = -1$是原分式方程的解，
∴$a = x = -1$，$b = \frac{x^{2}}{x + 2} = \frac{(-1)^{2}}{-1 + 2} = 1$，
∴$a + b = -1 + 1 = 0$．

15. (石家庄期末)解分式方程$\frac{k}{x - 2} = \frac{k - 2}{2 - x} - 3$产生增根，则$k$的值为1.

答案： 15．$1$ [解析]方程两边同乘$x - 2$，得$k = 2 - k - 3(x - 2)$．解这个整式方程，得$x = \frac{8 - 2k}{3}$．
∵分式方程$\frac{k}{x - 2} = 3$产生增根，
∴$\frac{8 - 2k}{3} = 2$，
∴$k = 1$．

16. (唐山期中)若$\frac{1}{a} - \frac{1}{b} = 4$，计算下列各式的值.
(1)$\frac{ab}{a - b} =$-\frac{1}{4}.
(2)$\frac{a - b}{3a + 2ab - 3b} =$\frac{2}{5}.

答案： 16．
(1)$-\frac{1}{4}$
(2)$\frac{2}{5}$ [解析]
(1)
∵$\frac{1}{a} - \frac{1}{b} = 4$，
∴$\frac{b - a}{ab} = 4$，即$b - a = 4ab$，整理，得$a - b = -4ab$，则原式$=\frac{ab}{-4ab} = -\frac{1}{4}$．
(2)
∵$a - b = -4ab$，
∴原式$=\frac{3(a - b) + 2ab}{-12ab + 2ab} = \frac{2}{5}$．

17. （8分）（唐山期末）计算.
(1)$\frac{x^{2}}{x - 2} - x - 2$
(2)$(\frac{12x}{x - 3} + x - 3) ÷ \frac{x^{2} + 6x + 9}{3x^{2} - 9x}$

答案： $(1)$计算$\frac{x^{2}}{x - 2} - x - 2$
解：
$\begin{aligned}&\frac{x^{2}}{x - 2} - x - 2\\=&\frac{x^{2}}{x - 2} - (x + 2)\\=&\frac{x^{2}}{x - 2} - \frac{(x + 2)(x - 2)}{x - 2}\\=&\frac{x^{2}}{x - 2} - \frac{x^{2}-4}{x - 2}\\=&\frac{x^{2}-(x^{2}-4)}{x - 2}\\=&\frac{x^{2}-x^{2}+4}{x - 2}\\=&\frac{4}{x - 2}\end{aligned}$
$(2)$计算$(\frac{12x}{x - 3} + x - 3) ÷ \frac{x^{2} + 6x + 9}{3x^{2} - 9x}$
解：
先对$\frac{12x}{x - 3} + x - 3$进行通分：
$\begin{aligned}&\frac{12x}{x - 3} + x - 3\\=&\frac{12x}{x - 3}+\frac{(x - 3)(x - 3)}{x - 3}\\=&\frac{12x+(x - 3)^{2}}{x - 3}\\=&\frac{12x+x^{2}-6x + 9}{x - 3}\\=&\frac{x^{2}+6x + 9}{x - 3}\end{aligned}$
再对$\frac{x^{2} + 6x + 9}{3x^{2} - 9x}$化简：$\frac{x^{2} + 6x + 9}{3x^{2} - 9x}=\frac{(x + 3)^{2}}{3x(x - 3)}$。
则原式$=\frac{x^{2}+6x + 9}{x - 3}÷\frac{(x + 3)^{2}}{3x(x - 3)}$
根据除法运算法则$a÷ b=a×\frac{1}{b}$，可得：
$\begin{aligned}&\frac{x^{2}+6x + 9}{x - 3}×\frac{3x(x - 3)}{(x + 3)^{2}}\\=&\frac{(x + 3)^{2}}{x - 3}×\frac{3x(x - 3)}{(x + 3)^{2}}\\=&3x\end{aligned}$
综上，答案依次为$(1)$$\boldsymbol{\frac{4}{x - 2}}$；$(2)$$\boldsymbol{3x}$。

18. （8分）（秦皇岛期中）解方程.
(1)$\frac{x}{x + 1} - 1 = \frac{3}{x - 1}$
(2)$\frac{3}{x + 2} + \frac{2}{x^{2} - 4} = \frac{1}{x - 2}$

答案： 18．解：
(1)方程两边同乘$(x + 1)(x - 1)$，得$x(x - 1) - (x + 1)(x - 1) = 3(x + 1)$，解这个方程，得$x = -\frac{1}{2}$．经检验，$x = -\frac{1}{2}$是原分式方程的解．
(2)方程两边同乘$(x + 2)(x - 2)$，得$3(x - 2) + 2 = x + 2$，解这个方程，得$x = 3$．经检验，$x = 3$是原分式方程的解．