《2025年名校课堂八年级数学上册人教版5四川专版》

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

答案： 解:原式$=\frac {2(x+2y)}{(x+2y)^{2}}=\frac {2}{x+2y}$.当$x=1$，$y=-2$时，原式$=\frac {2}{1+2×(-2)}=\frac {2}{1-4}=-\frac {2}{3}.$

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

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

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

答案： 解:$\because a:b:c=2:3:5$,
∴可设$a=2k$，$b=3k$，$c=5k$($k≠0$).
∴$\frac {b^{2}+c^{2}}{a^{2}}=\frac {(3k)^{2}+(5k)^{2}}{(2k)^{2}}=\frac {9k^{2}+25k^{2}}{4k^{2}}=\frac {34k^{2}}{4k^{2}}=\frac {17}{2}.$

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

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

【例】已知$x+\frac{1}{x}=3$，试求：
(1) $x^{2}+\frac{1}{x^{2}}$的值.
(2) $x-\frac{1}{x}$的值.
解：(1) $\because x+\frac{1}{x}=3$，
$\therefore(x+\frac{1}{x})^{2}=x^{2}+$2$+\frac{1}{x^{2}}=$9.
$\therefore x^{2}+\frac{1}{x^{2}}=$7.
(2) $\because x^{2}+\frac{1}{x^{2}}=$7，
$\therefore(x-\frac{1}{x})^{2}=$x^{2}-2+\frac{1}{x^{2}}$=$5.
$\therefore x-\frac{1}{x}=$\pm\sqrt{5}.
方法指导：(1) $x^{2}+\frac{1}{x^{2}}=(x\pm\frac{1}{x})^{2}\mp2$；
(2) $x^{4}+\frac{1}{x^{4}}=(x^{2}\pm\frac{1}{x^{2}})^{2}\mp2$；
(3) $(x+\frac{1}{x})^{2}-(x-\frac{1}{x})^{2}=4$.

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$x^{2}+\frac{1}{x^{2}}$的值为23.

答案： 23

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

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

答案： A

3. 已知$x+\frac{1}{x}=\frac{13}{6}$且$0\lt x\lt1$，求$x^{2}-\frac{1}{x^{2}}$的值.

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