8. 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.$26$
B.$24$
C.$22$
D.$20$

B

9. 比较大小：$\sqrt{5}-3\_\_\_\_\_\_\dfrac{\sqrt{5}-2}{2}$.(填“$＞$”“$＜$”或“$=$”)

＜

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.

$2\sqrt{3}+10\sqrt{2}$

11. 若$a$，$b$为有理数，且$\sqrt{8}+\sqrt{18}+\sqrt{\dfrac{1}{8}}=a + b\sqrt{2}$，则$a=$，$b=$.

0
$\frac{21}{4}$

12. 计算：
(1)$\sqrt{24}+\sqrt{\dfrac{1}{3}}-\sqrt{\dfrac{1}{27}}-\sqrt{36}$；
(2)$\sqrt{96}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{1}{8}}+\sqrt{6}$；
(3)$4b\sqrt{\dfrac{a}{b}}+\dfrac{2}{a}\sqrt{a^{3}b}-(3a\sqrt{\dfrac{b}{a}}+\sqrt{9ab})(a ＞ 0,b ＞ 0)$；
(4)$|\sqrt{2}-\sqrt{6}|+\sqrt{(\sqrt{2}-1)^{2}}-\sqrt{(\sqrt{6}-3)^{2}}$.

解：原式$​=2\sqrt {6}+\frac {\sqrt {3}}{3}-\frac {\sqrt {3}}{9}-6​$
$​=2\sqrt {6}+\frac {2}{9}\sqrt {3}-6​$
解：原式$​=4\sqrt {6}+\frac {\sqrt {2}}{2}-\frac {\sqrt {2}}{4}+\sqrt {6}​$
$​=5\sqrt {6}+\frac {\sqrt {2}}{4}​$
解：原式$​=4\sqrt {ab}+2\sqrt {ab}-(3\sqrt {ab}+3\sqrt {ab})​$
$​=6\sqrt {ab}-6\sqrt {ab}​$
​=0​
解：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：
$\dfrac{1}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1$；
$\dfrac{1}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}$；
$\dfrac{1}{\sqrt{4}+\sqrt{3}}=\dfrac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}=\sqrt{4}-\sqrt{3}$.

(1)观察上面的等式，请直接写出化简$\dfrac{1}{\sqrt{n + 1}+\sqrt{n}}$($n$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：$(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+···+\dfrac{1}{\sqrt{2026}+\sqrt{2025}})×(\sqrt{2026}+1)$.

$\sqrt{n+1}-\sqrt{n}$
1
解：原式$=(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\dots+\sqrt{2026}-\sqrt{2025})(\sqrt{2026}+1)$
$ =(\sqrt{2026}-1)(\sqrt{2026}+1)$
=2026-1=2025