《2025年名校课堂八年级数学上册人教版安徽专版》

【例】新考向 阅读理解 阅读理解：
因为$(a+\dfrac{1}{a})^{2}=a^{2}+2a\cdot \dfrac{1}{a}+(\dfrac{1}{a})^{2}=a^{2}+\dfrac{1}{a^{2}}+2$，①
$(a-\dfrac{1}{a})^{2}=a^{2}-2a\cdot \dfrac{1}{a}+(\dfrac{1}{a})^{2}=a^{2}+\dfrac{1}{a^{2}}-2$，②
所以由①可得$a^{2}+\dfrac{1}{a^{2}}=(a+\dfrac{1}{a})^{2}-2$，
由②可得$a^{2}+\dfrac{1}{a^{2}}=(a-\dfrac{1}{a})^{2}+2$.
同理$a^{4}+\dfrac{1}{a^{4}}=(a^{2}+\dfrac{1}{a^{2}})^{2}-2$.
试根据上面公式的变形解答下列问题：
(1)已知$a+\dfrac{1}{a}=2$，则下列等式成立的是C.
①$a^{2}+\dfrac{1}{a^{2}}=2$； ②$a^{4}+\dfrac{1}{a^{4}}=2$；
③$a-\dfrac{1}{a}=0$； ④$(a-\dfrac{1}{a})^{2}=2$.
A. ① B. ①②
C. ①②③ D. ①②③④
(2)已知$a+\dfrac{1}{a}=-2$，求下列代数式的值.
①$a^{2}+\dfrac{1}{a^{2}}$；②$(a-\dfrac{1}{a})^{2}$；③$a^{4}+\dfrac{1}{a^{4}}$.

答案：
(1)C；
(2)①原式$=(a+\dfrac{1}{a})^{2}-2=(-2)^{2}-2=2$；②原式$=a^{2}+\dfrac{1}{a^{2}}-2=2-2=0$；③原式$=(a^{2}+\dfrac{1}{a^{2}})^{2}-2=2^{2}-2=2$

1. 已知$x^{2}-3x+1=0$，则$x^{2}+\dfrac{1}{x^{2}}$的值为7.

答案： 7

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

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

3. 新考向 阅读理解 阅读理解：
【例】已知$x+\dfrac{1}{x}=3$，求分式$\dfrac{x}{x^{2}-4x+1}$的值.
解：因为$\dfrac{x^{2}-4x+1}{x}=x-4+\dfrac{1}{x}=x+\dfrac{1}{x}-4=3-4=-1$，所以$\dfrac{x}{x^{2}-4x+1}=-1$.
【活学活用】
(1)已知$a+\dfrac{1}{a}=-5$，求分式$\dfrac{2a^{2}+5a+2}{a}$的值.
(2)已知$b+\dfrac{1}{b}=-3$，求分式$\dfrac{b}{3b^{2}-4b+3}$的值.
(3)已知$x^{2}-4x-1=0$，求分式$\dfrac{3x^{2}}{x^{4}-7x^{2}+1}$的值.

答案：
(1)$\because a+\dfrac{1}{a}=-5,\therefore \dfrac{2a^{2}+5a+2}{a}=2a+5+\dfrac{2}{a}=2(a+\dfrac{1}{a})+5=2×(-5)+5=-5$；
(2)$\because b+\dfrac{1}{b}=-3,\therefore \dfrac{3b^{2}-4b+3}{b}=3b-4+\dfrac{3}{b}=3(b+\dfrac{1}{b})-4=3× (-3)-4=-13.\therefore \dfrac{b}{3b^{2}-4b+3}=-\dfrac{1}{13}$；
(3)$\because x^{2}-4x-1=0,\therefore x\neq0.\therefore x-4-\dfrac{1}{x}=0$，即$x-\dfrac{1}{x}=4.\therefore x^{2}-2+\dfrac{1}{x^{2}}=16$，即$x^{2}+\dfrac{1}{x^{2}}=18.\therefore \dfrac{x^{4}-7x^{2}+1}{3x^{2}}=\dfrac{1}{3}(x^{2}-7+\dfrac{1}{x^{2}})=\dfrac{1}{3}× (18-7)=\dfrac{11}{3}.\therefore \dfrac{3x^{2}}{x^{4}-7x^{2}+1}=\dfrac{3}{11}$