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9.（2023·海门区期末）如图，从一个大正方形中截去面积分别为8和18的两个小正方形，则图中阴影部分的面积为（ ）
A. 20
B. 22
C. 24
D. 26
A. 20
B. 22
C. 24
D. 26
答案
C
10. 已知等腰三角形的两边长分别为$2\sqrt{3}$和$5\sqrt{2}$，则此等腰三角形的周长为_______.
答案
$2\sqrt{3}+10\sqrt{2}$
11. 若$a,b$为有理数，且$\sqrt{8}+\sqrt{18}+\sqrt{\frac{1}{8}}=a + b\sqrt{2}$，则$a=$_______，$b=$_______.
答案
0 $\frac{21}{4}$
12. 计算：
(1)$\sqrt{24}+\sqrt{\frac{1}{3}}-(\sqrt{\frac{1}{27}}+6)$； (2)$\sqrt{96}+\sqrt{0.5}-(\sqrt{\frac{1}{8}}-\sqrt{6})$；
(3)$4b\sqrt{\frac{a}{b}}+\frac{2}{a}\sqrt{a^{3}b}-(3a\sqrt{\frac{b}{a}}+\sqrt{9ab})$； (4)$\vert\sqrt{2}-\sqrt{6}\vert+\sqrt{(\sqrt{2}-1)^{2}}-\sqrt{(\sqrt{6}-3)^{2}}$.
(1)$\sqrt{24}+\sqrt{\frac{1}{3}}-(\sqrt{\frac{1}{27}}+6)$； (2)$\sqrt{96}+\sqrt{0.5}-(\sqrt{\frac{1}{8}}-\sqrt{6})$；
(3)$4b\sqrt{\frac{a}{b}}+\frac{2}{a}\sqrt{a^{3}b}-(3a\sqrt{\frac{b}{a}}+\sqrt{9ab})$； (4)$\vert\sqrt{2}-\sqrt{6}\vert+\sqrt{(\sqrt{2}-1)^{2}}-\sqrt{(\sqrt{6}-3)^{2}}$.
答案
解：(1)原式$=2\sqrt{6}+\frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{9}-6=2\sqrt{6}+\frac{2}{9}\sqrt{3}-6$.
(2)原式$=4\sqrt{6}+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}+\sqrt{6}=5\sqrt{6}+\frac{\sqrt{2}}{4}$.
(3)原式$=4\sqrt{ab}+2\sqrt{ab}-(3\sqrt{ab}+3\sqrt{ab})=6\sqrt{ab}-6\sqrt{ab}=0$.
(4)原式$=\sqrt{6}-\sqrt{2}+\sqrt{2}-1-3+\sqrt{6}=2\sqrt{6}-4$.
(2)原式$=4\sqrt{6}+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}+\sqrt{6}=5\sqrt{6}+\frac{\sqrt{2}}{4}$.
(3)原式$=4\sqrt{ab}+2\sqrt{ab}-(3\sqrt{ab}+3\sqrt{ab})=6\sqrt{ab}-6\sqrt{ab}=0$.
(4)原式$=\sqrt{6}-\sqrt{2}+\sqrt{2}-1-3+\sqrt{6}=2\sqrt{6}-4$.
13. 阅读下面的材料，并解答问题：
$\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1$；
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}$；
$\frac{1}{\sqrt{4}+\sqrt{3}}=\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}=\sqrt{4}-\sqrt{3}$.
(1)观察上面的等式，请直接写出化简$\frac{1}{\sqrt{n + 1}+\sqrt{n}}$（$n$为正整数）的结果为______________；
(2)计算：$(\sqrt{n + 1}+\sqrt{n})(\sqrt{n + 1}-\sqrt{n})=$_______；
(3)请利用上面的规律及解法计算：$(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\cdots+\frac{1}{\sqrt{2025}+\sqrt{2024}})\times(\sqrt{2025}+1)$.
$\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1$；
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}$；
$\frac{1}{\sqrt{4}+\sqrt{3}}=\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}=\sqrt{4}-\sqrt{3}$.
(1)观察上面的等式，请直接写出化简$\frac{1}{\sqrt{n + 1}+\sqrt{n}}$（$n$为正整数）的结果为______________；
(2)计算：$(\sqrt{n + 1}+\sqrt{n})(\sqrt{n + 1}-\sqrt{n})=$_______；
(3)请利用上面的规律及解法计算：$(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\cdots+\frac{1}{\sqrt{2025}+\sqrt{2024}})\times(\sqrt{2025}+1)$.
答案
(1)$\sqrt{n + 1}-\sqrt{n}$ (2) 1
(3)解：原式$=(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\cdots+\sqrt{2025}-\sqrt{2024})(\sqrt{2025}+1)=(\sqrt{2025}-1)(\sqrt{2025}+1)=2025 - 1=2024$.
(3)解：原式$=(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\cdots+\sqrt{2025}-\sqrt{2024})(\sqrt{2025}+1)=(\sqrt{2025}-1)(\sqrt{2025}+1)=2025 - 1=2024$.
