9. 计算：$(\sqrt{3}+\sqrt{2})^{2}-\sqrt{24}=$；$\frac{\sqrt{3}}{\sqrt{3}+\sqrt{12}}=$.

9. 5 $ \frac{1}{3} $

$(\sqrt{3}+\sqrt{2})^{2}-\sqrt{24}$
$=(\sqrt{3})^{2}+2\sqrt{3}×\sqrt{2}+(\sqrt{2})^{2}-2\sqrt{6}$
$=3 + 2\sqrt{6} + 2 - 2\sqrt{6}$
$=5$
$\frac{\sqrt{3}}{\sqrt{3}+\sqrt{12}}$
$=\frac{\sqrt{3}}{\sqrt{3}+2\sqrt{3}}$
$=\frac{\sqrt{3}}{3\sqrt{3}}$
$=\frac{1}{3}$

10. 如果一个直角三角形的两条直角边长分别为$(\sqrt{6}+\sqrt{3})\mathrm{cm}$，$(3\sqrt{2}-3)\mathrm{cm}$，那么这个直角三角形的面积为$\mathrm{cm}^{2}$.

10. $ \frac{3\sqrt{3}}{2} $

直角三角形面积为两直角边乘积的一半，即：
$\begin{aligned}S&=\frac{1}{2}×(\sqrt{6}+\sqrt{3})×(3\sqrt{2}-3)\\&=\frac{1}{2}×[\sqrt{6}×3\sqrt{2}+\sqrt{6}×(-3)+\sqrt{3}×3\sqrt{2}+\sqrt{3}×(-3)]\\&=\frac{1}{2}×[3\sqrt{12}-3\sqrt{6}+3\sqrt{6}-3\sqrt{3}]\\&=\frac{1}{2}×(6\sqrt{3}-3\sqrt{3})\\&=\frac{1}{2}×3\sqrt{3}\\&=\frac{3\sqrt{3}}{2}\end{aligned}$
$\frac{3\sqrt{3}}{2}$

11. 如果$a+b=2+\sqrt{3}$，$ab=2\sqrt{3}$，那么$a-b$的值为.

11. $ 2 - \sqrt{3} $ 或 $ -2 + \sqrt{3} $

$(a-b)^2=(a+b)^2-4ab=(2+\sqrt{3})^2-4×2\sqrt{3}=4 + 4\sqrt{3} + 3 - 8\sqrt{3}=7 - 4\sqrt{3}$，
$a - b = \pm\sqrt{7 - 4\sqrt{3}} = \pm(2 - \sqrt{3})$，
即$a - b = 2 - \sqrt{3}$或$a - b = -2 + \sqrt{3}$。

12. 计算：
(1) $(\frac{\sqrt{5}-2}{3})^{2}$；
(2) $(5\sqrt{48}+\sqrt{12}-7\sqrt{7})÷\sqrt{3}$；
(3) $2\sqrt{12}×(3\sqrt{48}-\frac{4}{3}\sqrt{\frac{1}{8}}-3\sqrt{27})$；
(4) $(2\sqrt{5}+5\sqrt{2})×(2\sqrt{5}-5\sqrt{2})-(\sqrt{5}-\sqrt{2})^{2}$.

12. (1) $ \frac{9 - 4\sqrt{5}}{9} $ (2) $ 22 - \frac{7\sqrt{21}}{3} $ (3) $ 36 - \frac{4\sqrt{6}}{3} $ (4) $ 2\sqrt{10} - 37 $

(1) $(\frac{\sqrt{5}-2}{3})^{2}=\frac{(\sqrt{5})^{2}-2×\sqrt{5}×2+2^{2}}{3^{2}}=\frac{5 - 4\sqrt{5}+4}{9}=\frac{9 - 4\sqrt{5}}{9}$
(2) $(5\sqrt{48}+\sqrt{12}-7\sqrt{7})÷\sqrt{3}=5\sqrt{48÷3}+\sqrt{12÷3}-7\sqrt{7÷3}=5\sqrt{16}+\sqrt{4}-7×\frac{\sqrt{21}}{3}=5×4 + 2-\frac{7\sqrt{21}}{3}=20 + 2-\frac{7\sqrt{21}}{3}=22-\frac{7\sqrt{21}}{3}$
(3) $2\sqrt{12}×(3\sqrt{48}-\frac{4}{3}\sqrt{\frac{1}{8}}-3\sqrt{27})=4\sqrt{3}×(12\sqrt{3}-\frac{4}{3}×\frac{\sqrt{2}}{4}-9\sqrt{3})=4\sqrt{3}×(3\sqrt{3}-\frac{\sqrt{2}}{3})=4\sqrt{3}×3\sqrt{3}-4\sqrt{3}×\frac{\sqrt{2}}{3}=12×3-\frac{4\sqrt{6}}{3}=36-\frac{4\sqrt{6}}{3}$
(4) $(2\sqrt{5}+5\sqrt{2})×(2\sqrt{5}-5\sqrt{2})-(\sqrt{5}-\sqrt{2})^{2}=(2\sqrt{5})^{2}-(5\sqrt{2})^{2}-[(\sqrt{5})^{2}-2×\sqrt{5}×\sqrt{2}+(\sqrt{2})^{2}]=20 - 50-(5 - 2\sqrt{10}+2)=-30 - 7 + 2\sqrt{10}=2\sqrt{10}-37$

13. (2025·宿城段考)先化简，再求值：$(a+2b+\frac{3b^{2}}{a-2b})÷\frac{(a+b)^{2}}{2b-a}$，其中$a=\sqrt{2}$，$b=\sqrt{3}$.

13. 原式 $ = \frac{a^{2} - 4b^{2} + 3b^{2}}{a - 2b} · \frac{2b - a}{(a + b)^{2}} = \frac{a^{2} - b^{2}}{a - 2b} · \frac{2b - a}{(a + b)^{2}} = \frac{(a + b)(a - b)}{a - 2b} · \frac{2b - a}{(a + b)^{2}} = \frac{b - a}{b + a} $。$ \because a = \sqrt{2} $，$ b = \sqrt{3} $，$ \therefore $ 原式 $ = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} - \sqrt{2})^{2}}{(\sqrt{3} + \sqrt{2}) × (\sqrt{3} - \sqrt{2})} = 5 - 2\sqrt{6} $

14. 从$-\sqrt{2}$，$\sqrt{3}$，$\sqrt{6}$中任意选择两个数，分别填在算式$(□+◯)^{2}÷\sqrt{2}$里面的“$□$”与“$◯$”中，并计算该算式的结果.

14. 若选择的数是 $ -\sqrt{2} $ 和 $ \sqrt{3} $，则 $ (-\sqrt{2} + \sqrt{3})^{2} ÷ \sqrt{2} = (5 - 2\sqrt{6}) ÷ \sqrt{2} = \frac{5\sqrt{2}}{2} - 2\sqrt{3} $；若选择的数是 $ -\sqrt{2} $ 和 $ \sqrt{6} $，则 $ (-\sqrt{2} + \sqrt{6})^{2} ÷ \sqrt{2} = (8 - 2\sqrt{12}) ÷ \sqrt{2} = 4\sqrt{2} - 2\sqrt{6} $；若选择的数是 $ \sqrt{3} $ 和 $ \sqrt{6} $，则 $ (\sqrt{3} + \sqrt{6})^{2} ÷ \sqrt{2} = (9 + 2\sqrt{18}) ÷ \sqrt{2} = \frac{9\sqrt{2}}{2} + 6 $