《2025年考出好成绩八年级数学上册青岛版山东专版》

9. 等式$\frac {8x + 9}{(x + 3)(x - 2)} = \frac {A}{x + 3} + \frac {B}{x - 2}$对于任何使分母不为 0 的$x$均成立,求$A$,$B$的值。

答案： 解：$\because \frac {A}{x+3}+\frac {B}{x-2}$$=\frac {A(x-2)+B(x+3)}{(x+3)(x-2)}$$=\frac {(A+B)x+(3B-2A)}{(x+3)(x-2)}$$=\frac {8x+9}{(x+3)(x-2)}.$$\therefore \left\{\begin{array}{l} A+B=8,\\ 3B-2A=9,\end{array}\right. $解得$\left\{\begin{array}{l} A=3,\\ B=5.\end{array}\right. $

10. 提供方法支架 阅读下列解题过程：
已知$\frac {x}{x^{2} + 1} = \frac {1}{2}$,求$\frac {x^{2}}{x^{4} + 1}$的值。
解:由$\frac {x}{x^{2} + 1} = \frac {1}{2}$知,$x \neq 0$,
$\therefore \frac {x^{2} + 1}{x} = 2$,即$x + \frac {1}{x} = 2$,
$\therefore \frac {x^{4} + 1}{x^{2}} = x^{2} + \frac {1}{x^{2}} = (x + \frac {1}{x})^{2} - 2 = 2^{2} - 2 = 2$,
$\therefore \frac {x^{2}}{x^{4} + 1}$的值为 2 的倒数,即$\frac {1}{2}$。
以上解法中先将已知等式的两边“取倒数”,然后求出待求式子倒数的值,我们把这种解法叫作“倒数法”,请你利用“倒数法”解决下列问题：
(1) 已知$\frac {x}{x^{2} + 1} = \frac {1}{3}$,求$\frac {x^{2}}{x^{4} + 1}$的值；
(2) 已知$\frac {x}{x^{2} - x + 1} = \frac {1}{4}$,求$\frac {x^{2}}{x^{4} - 2x^{2} + 1}$的值；
(3) 已知$\frac {xy}{x + y} = 2$,$\frac {yz}{y + z} = \frac {4}{3}$,$\frac {zx}{z + x} = \frac {4}{3}$,求$\frac {xyz}{xy + yz + zx}$的值。

答案： 解：
(1)由$\frac {x}{x^{2}+1}=\frac {1}{3}$知，$x≠0,$$\therefore \frac {x^{2}+1}{x}=3$,即$x+\frac {1}{x}=3,$$\therefore \frac {x^{4}+1}{x^{2}}=x^{2}+\frac {1}{x^{2}}=(x+\frac {1}{x})^{2}-2=3^{2}-2=7,$$\therefore \frac {x^{2}}{x^{4}+1}$的值为7的倒数，即$\frac {1}{7}.$
(2)由$\frac {x}{x^{2}-x+1}=\frac {1}{4}$知，$x≠0,$$\therefore \frac {x^{2}-x+1}{x}=4$,即$x+\frac {1}{x}-1=4,$$\therefore x+\frac {1}{x}=5,$$\therefore \frac {x^{4}-2x^{2}+1}{x^{2}}=x^{2}+\frac {1}{x^{2}}-2=(x+\frac {1}{x})^{2}-2-2=5^{2}-2$$-2=21.$$\therefore \frac {x^{2}}{x^{4}-2x^{2}+1}$的值为21的倒数，即$\frac {1}{21}.$
(3)由$\frac {xy}{x+y}=2,\frac {yz}{y+z}=\frac {4}{3},\frac {zx}{z+x}=\frac {4}{3}$知，$x≠0,y≠0,z$$≠0,$$\therefore \frac {x+y}{xy}=\frac {1}{x}+\frac {1}{y}=\frac {1}{2},\frac {y+z}{yz}=\frac {1}{y}+\frac {1}{z}=\frac {3}{4},\frac {z+x}{zx}=\frac {1}{x}$$+\frac {1}{z}=\frac {3}{4},$$\therefore \frac {1}{x}+\frac {1}{y}+\frac {1}{z}=1.$$\therefore \frac {xy+yz+zx}{xyz}=\frac {1}{x}+\frac {1}{y}+\frac {1}{z},$$\therefore \frac {xyz}{xy+yz+zx}=\frac {1}{\frac {1}{x}+\frac {1}{y}+\frac {1}{z}}=1.$