《2025年1加1轻巧夺冠完美期末八年级数学上册人教版辽宁专版》

22. (12分)
阅读下面的解题过程:
已知$\frac{x}{x^{2}+1}=\frac{1}{3}$,求$\frac{x^{2}}{x^{4}+1}$的值.
解:由$\frac{x}{x^{2}+1}=\frac{1}{3}$知$x\neq0$,所以$\frac{x^{2}+1}{x}=3$,即$x+\frac{1}{x}=3$,所以$\frac{x^{4}+1}{x^{2}}=x^{2}+\frac{1}{x^{2}}=(x+\frac{1}{x})^{2}-2=3^{2}-2=7$,故$\frac{x^{2}}{x^{4}+1}$的值为$\frac{1}{7}$.
上题的解法叫作“倒数法”,请你利用“倒数法”解决下面的题目:
(1)已知$\frac{x}{x^{2}-3x+1}=-1$,求$\frac{x^{2}}{x^{4}-7x^{2}+1}$的值.
(2)已知$\frac{1}{a}+\frac{1}{b}=\frac{1}{6}$,$\frac{1}{b}+\frac{1}{c}=\frac{1}{9}$,$\frac{1}{a}+\frac{1}{c}=\frac{1}{15}$,求$\frac{abc}{ab+bc+ac}$的值.

答案： 22.解:
(1)由$\frac{x}{x^2 - 3x + 1}$ = - 1知x≠0，所以$\frac{x^2 - 3x + 1}{x}$ = - 1，即x + $\frac{1}{x}$ = 2，所以$\frac{x^4 - 7x^2 + 1}{x^2}$ = $x^2 + \frac{1}{x^2} - 7$ = $(x + \frac{1}{x})^2 - 2 - 7$ = $2^2 - 2 - 7$ = - 5，故$\frac{x^2}{x^4 - 7x^2 + 1}$ = - $\frac{1}{5}$.
(2)由题知2$(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ = $\frac{1}{6} + \frac{1}{9} + \frac{1}{15}$ = $\frac{31}{90}$，所以$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ = $\frac{31}{180}$，所以$\frac{ab + bc + ac}{abc}$ = $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$，故$\frac{abc}{ab + bc + ac}$ = $\frac{180}{31}$.