《2025年单元双测全优测评卷八年级数学下册沪科版》

19.已知a²+b²−4a−2b+5=0,求$\frac{\sqrt{a}−b}{2\sqrt{a}−b−1}$的值.

答案： 将已知等式变形为$(a - 2)^2+(b - 1)^2 = 0$，
所以$a - 2 = 0$，$b - 1 = 0$，
解得$a = 2$，$b = 1$，
所以$\frac{\sqrt{a}-b}{2\sqrt{a}-b - 1}=\frac{\sqrt{2}-1}{2\sqrt{2}-1 - 1}=\frac{1}{2}$.

20.阅读下面的问题:
$\frac{1}{1+\sqrt{2}}$=$\frac{1×(\sqrt{2}−1)}{\sqrt{2}+1)(\sqrt{2}−1)}$=$\sqrt{2}$−1;
$\frac{1}{3+\sqrt{2}}$=$\frac{\sqrt{3}\sqrt{2}}{\sqrt{3}+\sqrt{2})(\sqrt{3}−\sqrt{2})}$=$\sqrt{3}$-$\sqrt{2}$;
$\frac{1}{5+2}$=$\frac{\sqrt{5}−2}{\sqrt{5}+2)(\sqrt{5}−2)}$=$\sqrt{5}$−2;
试求:(1)$\frac{1}{7+\sqrt{6}}$;(2)$\frac{1}{3\sqrt{2}+\sqrt{17}}$;(3)$\frac{1}{n+1+\sqrt{n}}$(n为正整数).

答案：
(1)$\sqrt{7}-\sqrt{6}$
(2)$3\sqrt{2}-\sqrt{17}$
(3)$\sqrt{n + 1}-\sqrt{n}$