8. 计算：$(\sqrt{3}+\sqrt{2})^{2}-\sqrt{24}=$
$\frac{1}{3}$
；$\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{12}}=$
$\frac{1}{3}$
.
答案：$8. \frac{1}{3}$
解析：
$(\sqrt{3}+\sqrt{2})^{2}-\sqrt{24}$
$=3 + 2\sqrt{6} + 2 - 2\sqrt{6}$
$=5$
$\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{12}}$
$=\dfrac{\sqrt{3}}{\sqrt{3}+2\sqrt{3}}$
$=\dfrac{\sqrt{3}}{3\sqrt{3}}$
$=\dfrac{1}{3}$
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9. 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
$\frac{3\sqrt{3}}{2}$
$\mathrm{cm}^{2}$.
答案：$9. \frac{3\sqrt{3}}{2}$
解析：
$\frac{1}{2} × (\sqrt{6} + \sqrt{3}) × (3\sqrt{2} - 3)$
$=\frac{1}{2} × [\sqrt{6} × 3\sqrt{2} - \sqrt{6} × 3 + \sqrt{3} × 3\sqrt{2} - \sqrt{3} × 3]$
$=\frac{1}{2} × [3\sqrt{12} - 3\sqrt{6} + 3\sqrt{6} - 3\sqrt{3}]$
$=\frac{1}{2} × (6\sqrt{3} - 3\sqrt{3})$
$=\frac{1}{2} × 3\sqrt{3}$
$=\frac{3\sqrt{3}}{2}$
10. 如果$a + b = 2+\sqrt{3}$，$ab = 2\sqrt{3}$，那么$a - b$的值为
$2 - \sqrt{3}$或$ - 2 + \sqrt{3}$
.
答案：$10. 2 - \sqrt{3}$或$ - 2 + \sqrt{3}$
解析：
$(a - b)^2 = (a + b)^2 - 4ab$
$=(2 + \sqrt{3})^2 - 4×2\sqrt{3}$
$=4 + 4\sqrt{3} + 3 - 8\sqrt{3}$
$=7 - 4\sqrt{3}$
$a - b = \pm\sqrt{7 - 4\sqrt{3}} = \pm(2 - \sqrt{3})$
即$a - b = 2 - \sqrt{3}$或$a - b = -2 + \sqrt{3}$
11. 计算：
（1）$2\sqrt{12}×(3\sqrt{48}-\dfrac{4}{3}\sqrt{\dfrac{1}{8}}-3\sqrt{27})$；
（2）$(5\sqrt{48}+\sqrt{12}-7\sqrt{7})÷\sqrt{3}$；
（3）$(\dfrac{\sqrt{5}-2}{3})^{2}$；
（4）$(2\sqrt{5}+5\sqrt{2})×(2\sqrt{5}-5\sqrt{2})-(\sqrt{5}-\sqrt{2})^{2}$.
答案：$11. (1) 36 - \frac{4\sqrt{6}}{3} (2) 22 - \frac{7\sqrt{21}}{3} (3) \frac{9 - 4\sqrt{5}}{9} (4) 2\sqrt{10} - 37$
解析：
（1）$2\sqrt{12}×(3\sqrt{48}-\dfrac{4}{3}\sqrt{\dfrac{1}{8}}-3\sqrt{27})$
$=2×2\sqrt{3}×(3×4\sqrt{3}-\dfrac{4}{3}×\dfrac{\sqrt{2}}{4}-3×3\sqrt{3})$
$=4\sqrt{3}×(12\sqrt{3}-\dfrac{\sqrt{2}}{3}-9\sqrt{3})$
$=4\sqrt{3}×(3\sqrt{3}-\dfrac{\sqrt{2}}{3})$
$=4\sqrt{3}×3\sqrt{3}-4\sqrt{3}×\dfrac{\sqrt{2}}{3}$
$=36 - \dfrac{4\sqrt{6}}{3}$
（2）$(5\sqrt{48}+\sqrt{12}-7\sqrt{7})÷\sqrt{3}$
$=(5×4\sqrt{3}+2\sqrt{3}-7\sqrt{7})÷\sqrt{3}$
$=(20\sqrt{3}+2\sqrt{3}-7\sqrt{7})÷\sqrt{3}$
$=(22\sqrt{3}-7\sqrt{7})÷\sqrt{3}$
$=22\sqrt{3}÷\sqrt{3}-7\sqrt{7}÷\sqrt{3}$
$=22 - \dfrac{7\sqrt{21}}{3}$
（3）$(\dfrac{\sqrt{5}-2}{3})^{2}$
$=\dfrac{(\sqrt{5})^{2}-2×\sqrt{5}×2 + 2^{2}}{3^{2}}$
$=\dfrac{5 - 4\sqrt{5} + 4}{9}$
$=\dfrac{9 - 4\sqrt{5}}{9}$
（4）$(2\sqrt{5}+5\sqrt{2})×(2\sqrt{5}-5\sqrt{2})-(\sqrt{5}-\sqrt{2})^{2}$
$=(2\sqrt{5})^{2}-(5\sqrt{2})^{2}-[(\sqrt{5})^{2}-2×\sqrt{5}×\sqrt{2}+(\sqrt{2})^{2}]$
$=20 - 50 - (5 - 2\sqrt{10} + 2)$
$=-30 - (7 - 2\sqrt{10})$
$=-30 - 7 + 2\sqrt{10}$
$=2\sqrt{10} - 37$
12. $(2025·广州)$求代数式$\dfrac{2m^{2}+4m}{m - 2}·\dfrac{m^{2}-4m + 4}{m}$的值，其中$m=\sqrt{3}-1$.
答案：12. 原式$ = \frac{2m(m + 2)}{m - 2} · \frac{(m - 2)^2}{m} = 2(m + 2)(m - 2) = 2(m^2 - 4) = 2m^2 - 8. $当$ m = \sqrt{3} - 1 $时，原式$ = 2(\sqrt{3} - 1)^2 - 8 = 2(3 - 2\sqrt{3} + 1) - 8 = 8 - 4\sqrt{3} - 8 = - 4\sqrt{3}$
13. 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.
答案：13. 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$
