《2025年名校课堂八年级数学上册人教版云南专版》

1. 先约分，再求值：$\frac{2x + 4y}{x^2 + 4xy + 4y^2}$，其中$x = 1$，$y = -2$。

答案： 1. 解：原式$ = \frac {2(x + 2y)}{(x + 2y)^2} = \frac {2}{x + 2y}$. 当 $x = 1$, $y = -2$, 原式$ = \frac {2}{1 + 2 × (-2)} = - \frac {2}{3}$.

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

答案： 2. 解：$\because \frac {x}{y} = 2$, $\therefore x = 2y$. $\therefore$原式$ = \frac {(x + y)(x - y)}{(x - y)^2} = \frac {x + y}{x - y} = \frac {3y}{y} = 3$.

3. 已知$a : b : c = 2 : 3 : 5$，求$\frac{b^2 + c^2}{a^2}$的值。

答案： 3. 解：$\because a : b : c = 2 : 3 : 5$, $\therefore$可设 $a = 2k$, $b = 3k$, $c = 5k (k \neq 0)$. $\therefore \frac {b^2 + c^2}{a^2} = \frac {9k^2 + 25k^2}{4k^2} = \frac {17}{2}$.

4. 已知$m^2 - 3m + 2 = 0$，求$\frac{2025m}{m^2 - 2m + 2}$的值。

答案： 4. 解：$\because m^2 - 3m + 2 = 0$, $\therefore m^2 + 2 = 3m$. $\therefore$原式$ = \frac {2025m}{3m - 2m} = \frac {2025m}{m} = 2025$.

【例】已知 $ x + \dfrac{1}{x} = 3 $，试求：
(1) $ x^{2} + \dfrac{1}{x^{2}} $的值；
(2) $ x - \dfrac{1}{x} $的值.
解：(1) $ \because x + \dfrac{1}{x} = 3 $，
$ \therefore (x + \dfrac{1}{x})^{2} = x^{2} + $2$ + \dfrac{1}{x^{2}} = $9.
$ \therefore x^{2} + \dfrac{1}{x^{2}} = $7.
(2) $ \because x^{2} + \dfrac{1}{x^{2}} = $7，
$ \therefore (x - \dfrac{1}{x})^{2} = $$x^{2}-2+\frac{1}{x^{2}}$$ = $5.
$ \therefore x - \dfrac{1}{x} = $$\pm\sqrt{5}$.
方法指导：(1) $ x^{2} + \dfrac{1}{x^{2}} = (x \pm \dfrac{1}{x})^{2} \mp 2 $；
(2) $ x^{4} + \dfrac{1}{x^{4}} = (x^{2} \pm \dfrac{1}{x^{2}})^{2} \mp 2 $；
(3) $ (x + \dfrac{1}{x})^{2} - (x - \dfrac{1}{x})^{2} = 4 $.

答案：
(1)2 9 7
(2)7 $x^{2}-2+\frac{1}{x^{2}}$ 5 $\pm\sqrt{5}$

1. 已知 $ x^{2} - 5x + 1 = 0 $，则 $ x^{2} + \dfrac{1}{x^{2}} $的值为23.

答案： 1.23

2. (教材 P145 新增习题 T12 变式)若 $ x - \dfrac{1}{x} = 4 $，则 $ \dfrac{3x^{2}}{x^{4} - 7x^{2} + 1} = $ (A)

A.$ \dfrac{3}{11} $
B.$ -1 $
C.$ \dfrac{1}{3} $
D.$ -\dfrac{3}{5} $

答案： 2.A

3. 已知 $ x + \dfrac{1}{x} = \dfrac{13}{6} $且 $ 0 ＜ x ＜ 1 $，求 $ x^{2} - \dfrac{1}{x^{2}} $的值.

答案： 3.解：$\because x+\frac{1}{x}=\frac{13}{6}$,$\therefore (x+\frac{1}{x})^{2}=\frac{169}{36}$,$\therefore x^{2}+2+\frac{1}{x^{2}}=\frac{169}{36}$,$\therefore x^{2}-2+\frac{1}{x^{2}}=\frac{25}{36}$,$\therefore (x-\frac{1}{x})^{2}=\frac{25}{36}$,$\because 0＜x＜1$,$\therefore x-\frac{1}{x}＜0$,$\therefore x-\frac{1}{x}=$
$-\frac{5}{6}$,$\therefore x^{2}-\frac{1}{x^{2}}=(x+\frac{1}{x})(x-\frac{1}{x})=\frac{13}{6}×(-\frac{5}{6})=-\frac{65}{36}$.