《2025年千里马单元测试卷八年级数学下册人教版》

1. 已知$a = \frac{1}{\sqrt{2}-1}$，求$2a^{2}-4a + 1$的值.

答案： 解：$\because a=\frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{2}+1$，
$\therefore a - 1=\sqrt{2}$，$\therefore (a - 1)^2 = 2$，$a^2 - 2a + 1 = 2$，
$\therefore a^2 - 2a = 1$，$\therefore 2a^2 - 4a = 2$，
$\therefore 2a^2 - 4a + 1 = 3$.

2. (齐齐哈尔龙沙区期中)化简:$\frac{a\sqrt{a}-b\sqrt{a}}{a+\sqrt{ab}}$.

答案： 解：原式$=\frac{\sqrt{a}(a - b)}{\sqrt{a}(\sqrt{a}+\sqrt{b})}=\sqrt{a}-\sqrt{b}$.

3. 已知$a + b = -2\sqrt{2},ab = 1$，求$\sqrt{\frac{2b}{a}}+\sqrt{\frac{2a}{b}}$的值.

答案： 解：$\because (\sqrt{\frac{2b}{a}}+\sqrt{\frac{2a}{b}})^2=\frac{2b}{a}+\frac{2a}{b}+4=\frac{2(a + b)^2}{ab}$，
把$a + b=-2\sqrt{2}$，$ab = 1$代入，得$(\sqrt{\frac{2b}{a}}+\sqrt{\frac{2a}{b}})^2 = 16$，
$\therefore \sqrt{\frac{2b}{a}}+\sqrt{\frac{2a}{b}} = 4$.

4. 若$\sqrt{27 - a^{2}}+\sqrt{9 + a^{2}} = 7$，求$\sqrt{27 - a^{2}}-\sqrt{9 + a^{2}}$的值.

答案： 解：设$\sqrt{27 - a^2}=x$，$\sqrt{9 + a^2}=y$，
则$x + y = 7$，
$\therefore x^2 + y^2 = 27 - a^2 + 9 + a^2 = 36$.
$\because (x + y)^2 - 2xy = x^2 + y^2$，
即$49 - 2xy = 36$，
$\therefore 2xy = 13$，
$\therefore (x - y)^2 = x^2 + y^2 - 2xy = 36 - 13 = 23$，
$\therefore x - y=\pm\sqrt{23}$，
$\therefore$原式$=\pm\sqrt{23}$.

5. 已知$\sqrt{x}+\frac{1}{\sqrt{x}} = 3$，求$\frac{x}{x^{2}+3x + 1}$的值.

答案： 解：设$t=\frac{x}{x^2 + 3x + 1}$，则$\frac{1}{t}=\frac{x^2 + 3x + 1}{x}=x + 3+\frac{1}{x}$.
因为$\sqrt{x}+\frac{1}{\sqrt{x}} = 3$，所以$(\sqrt{x}+\frac{1}{\sqrt{x}})^2 = 3^2$，
即$x+\frac{1}{x}+2 = 9$，
所以$x+\frac{1}{x}=7$，所以$\frac{1}{t}=x + 3+\frac{1}{x}=7 + 3 = 10$，
所以$t=\frac{1}{10}$，即$\frac{x}{x^2 + 3x + 1}=\frac{1}{10}$.