《2025年通城学典课时作业本八年级数学下册苏科版江苏专版》

1. 已知$a、b$互为倒数，则代数式$\frac{a^{2}-2ab + b^{2}}{a - b}\div(\frac{1}{b}-\frac{1}{a})$的值为________.

答案： 1

2. （2024·遂宁）先化简：$(1 - \frac{1}{x - 1})\div\frac{x - 2}{x^{2}-2x + 1}$，再从1、2、3中选择一个合适的数作为$x$的值代入求值.

答案： 原式 = $x - 1$. 由题意，得 $x - 1 \neq 0$，$x - 2 \neq 0$，$x^{2} - 2x + 1 \neq 0$，$\therefore x \neq 1$，$x \neq 2$. $\therefore x$的值只能取3. 此时原式 = 2

3. 若$x+\frac{1}{x}=\frac{13}{6}$，且$0＜x＜1$，则$x^{2}-\frac{1}{x^{2}}$的值为________.

答案： $-\frac{65}{36}$ 解析：$\because 0 ＜ x ＜ 1$，$\therefore x ＜ \frac{1}{x}$. $\therefore x - \frac{1}{x} ＜ 0$. $\because x + \frac{1}{x} = \frac{13}{6}$，$\therefore (x + \frac{1}{x})^{2} = \frac{169}{36}$，即 $x^{2} + 2 + \frac{1}{x^{2}} = \frac{169}{36}$. $\therefore x^{2} - 2 + \frac{1}{x^{2}} = \frac{169}{36} - 4$. $\therefore (x - \frac{1}{x})^{2} = \frac{25}{36}$. $\therefore x - \frac{1}{x} = -\frac{5}{6}$. $\therefore x^{2} - \frac{1}{x^{2}} = (x + \frac{1}{x})(x - \frac{1}{x}) = \frac{13}{6} \times (-\frac{5}{6}) = -\frac{65}{36}$.

4. 已知$y＞x＞0$，$x^{2}+y^{2}=4xy$，求$\frac{x + y}{x - y}$的值.

答案： $\because x^{2} + y^{2} = 4xy$，$\therefore x^{2} + 2xy + y^{2} = 6xy$，$x^{2} - 2xy + y^{2} = 2xy$. $\therefore (x + y)^{2} = 6xy$，$(x - y)^{2} = 2xy$. $\therefore \frac{(x + y)^{2}}{(x - y)^{2}} = \frac{6xy}{2xy}$，即 $(\frac{x + y}{x - y})^{2} = 3$. $\because y ＞ x ＞ 0$，$\therefore \frac{x + y}{x - y} ＜ 0$. $\therefore \frac{x + y}{x - y} = -\sqrt{3}$

5. （2024·大庆）若$a+\frac{1}{a}=\sqrt{5}$，则$a^{2}+\frac{1}{a^{2}}=$________.

答案： 3

6. （2024·北京）先化简，再求值：$\frac{3(a - 2b)+3b}{a^{2}-2ab + b^{2}}$，其中$a、b$满足$a - b - 1 = 0$.

答案： 原式 = $\frac{3}{a - b}$. $\because a - b - 1 = 0$，$\therefore a - b = 1$. $\therefore$原式 = 3

7. 已知$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$，$\frac{1}{b}+\frac{1}{c}=\frac{1}{3}$，$\frac{1}{a}+\frac{1}{c}=\frac{1}{4}$，求$\frac{abc}{ab + bc + ac}$的值.

答案： $\because \frac{1}{a} + \frac{1}{b} = \frac{1}{2}$，$\frac{1}{b} + \frac{1}{c} = \frac{1}{3}$，$\frac{1}{a} + \frac{1}{c} = \frac{1}{4}$，$\therefore \frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{a} + \frac{1}{c} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$，即 $\frac{2}{a} + \frac{2}{b} + \frac{2}{c} = \frac{13}{12}$. $\therefore \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{13}{24}$. 根据题意，得 $abc \neq 0$，$\therefore \frac{abc}{ab + bc + ac} = \frac{abc \div abc}{(ab + bc + ac) \div abc} = \frac{1}{\frac{1}{c} + \frac{1}{a} + \frac{1}{b}} = \frac{24}{13}$