《2025年点石成金金牌夺冠八年级数学上册人教版辽宁专版》

1. 某超市用3000元购进某种干果销售，由于销售状况良好，超市又花费9000元购进该种干果，但这次的进价是第一次进价的1.5倍，购进干果的数量比第一次多300千克.这种干果第一次的进价是每千克多少元?

答案： 1.解：设这种干果第一次的进价是每千克$x$元，则第二次的进价是每千克$1.5x$元.
根据题意，得$\frac{3000}{x} + 300 = \frac{9000}{1.5x}$.
方程两边乘$1.5x$，
得$3000 × 1.5 + 300 × 1.5x = 9000$.
解得$x = 10$.
检验：当$x = 10$时，$1.5x \neq 0$.
$\therefore$原分式方程的解是$x = 10$.
答：这种干果第一次的进价是每千克10元.

2. 已知$\frac{x}{x^{2}+1}=\frac{1}{3}$，求$\frac{x^{2}}{x^{4}+1}$的值.
解：$\because\frac{x}{x^{2}+1}=\frac{1}{3}$，$\therefore\frac{x^{2}+1}{x}=3$，即$x+\frac{1}{x}=3$.
$\therefore\frac{x^{4}+1}{x^{2}}=x^{2}+\frac{1}{x^{2}}=(x+\frac{1}{x})^{2}-2=3^{2}-2=7$，$\therefore\frac{x^{2}}{x^{4}+1}=\frac{1}{7}$.
上面的解法叫作“倒数法”，请你利用“倒数法”解答下列问题：
(1)已知$\frac{x}{x^{2}-3x+1}=-1$，求$\frac{x^{2}}{x^{4}-7x^{2}+1}$的值.
(2)已知$\frac{ab}{a+b}=6$，$\frac{bc}{b+c}=9$，$\frac{ac}{a+c}=15$，求$\frac{abc}{ab+bc+ac}$的值.

答案： 2.解：
(1)$\because \frac{x}{x^2 - 3x + 1} = -1$，
$\therefore \frac{x^2 - 3x + 1}{x} = x + \frac{1}{x} - 3 = -1$.
$\therefore x + \frac{1}{x} = 2$，$\therefore (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 = 4$.
$\therefore x^2 + \frac{1}{x^2} = 2$.
$\therefore \frac{x^4 - 7x^2 + 1}{x^2} = x^2 + \frac{1}{x^2} - 7 = 2 - 7 = -5$，
$\therefore \frac{x^2}{x^4 - 7x^2 + 1} = \frac{1}{5}$.
(2)$\because \frac{ab}{a + b} = 6$，$\frac{bc}{b + c} = 9$，$\frac{ac}{a + c} = 15$，
$\therefore \frac{1}{a} + \frac{1}{b} = \frac{1}{6}$，$\frac{1}{b} + \frac{1}{c} = \frac{1}{9}$，$\frac{1}{a} + \frac{1}{c} = \frac{1}{15}$.
$\therefore 2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = \frac{1}{6} + \frac{1}{9} + \frac{1}{15} = \frac{31}{90}$.
$\therefore \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{31}{180}$.
$\therefore \frac{ab + bc + ac}{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{31}{180}$，
$\therefore \frac{abc}{ab + bc + ac} = \frac{180}{31}$.