三、解答题
10. 计算:
(1)$\sqrt{24}÷\sqrt{3}+\sqrt{6}×2\sqrt{3}$; (2)$\sqrt{45}+\sqrt{5}-3\sqrt{\frac{1}{3}}×\sqrt{27}$;
(3)$\sqrt{8}+\sqrt{12}-|\sqrt{2}-\sqrt{3}|$; (4)$(3-\sqrt{7})(3+\sqrt{7})+\sqrt{2}(2-\sqrt{2})$.
10. 计算:
(1)$\sqrt{24}÷\sqrt{3}+\sqrt{6}×2\sqrt{3}$; (2)$\sqrt{45}+\sqrt{5}-3\sqrt{\frac{1}{3}}×\sqrt{27}$;
(3)$\sqrt{8}+\sqrt{12}-|\sqrt{2}-\sqrt{3}|$; (4)$(3-\sqrt{7})(3+\sqrt{7})+\sqrt{2}(2-\sqrt{2})$.
答案：10.解:(1)原式$=2\sqrt{2}+6\sqrt{2}=8\sqrt{2}$.
(2)原式$=3\sqrt{5}+\sqrt{5}-3\sqrt{\frac{1}{3}×27}=4\sqrt{5}-9$.
(3)原式$=2\sqrt{2}+2\sqrt{3}+\sqrt{2}-\sqrt{3}=3\sqrt{2}+\sqrt{3}$.
(4)原式$=9-7+2\sqrt{2}-2=2\sqrt{2}$.
(2)原式$=3\sqrt{5}+\sqrt{5}-3\sqrt{\frac{1}{3}×27}=4\sqrt{5}-9$.
(3)原式$=2\sqrt{2}+2\sqrt{3}+\sqrt{2}-\sqrt{3}=3\sqrt{2}+\sqrt{3}$.
(4)原式$=9-7+2\sqrt{2}-2=2\sqrt{2}$.
11. (1)已知x,y是实数,且满足$y=\sqrt{x-2}+\sqrt{2-x}+\frac{1}{8}$,求$\sqrt{x}·\sqrt{y}$的值;
(2)比较数$3\sqrt{5}$和$2\sqrt{11}$的大小.
(2)比较数$3\sqrt{5}$和$2\sqrt{11}$的大小.
答案：11.解:(1)$\because y=\sqrt{x-2}+\sqrt{2-x}+\frac{1}{8}$,
$\therefore x-2≥0,2-x≥0,\therefore x=2,y=\frac{1}{8}$,
则原式$=\sqrt{2}×\sqrt{\frac{1}{8}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$.
(2)$\because 3\sqrt{5}=\sqrt{45},2\sqrt{11}=\sqrt{44},\sqrt{45}＞\sqrt{44}$,
$\therefore 3\sqrt{5}＞2\sqrt{11}$.
$\therefore x-2≥0,2-x≥0,\therefore x=2,y=\frac{1}{8}$,
则原式$=\sqrt{2}×\sqrt{\frac{1}{8}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$.
(2)$\because 3\sqrt{5}=\sqrt{45},2\sqrt{11}=\sqrt{44},\sqrt{45}＞\sqrt{44}$,
$\therefore 3\sqrt{5}＞2\sqrt{11}$.
12. 在解决问题“已知$a=\frac{1}{2+\sqrt{3}}$,求$2a^{2}-8a+1$的值”时,小明是这样分析与解答的:
$\because a=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3},\therefore a-2=-\sqrt{3}$,
$\therefore(a-2)^{2}=3$,即$a^{2}-4a+4=3,\therefore a^{2}-4a=-1$,
$\therefore2a^{2}-8a+1=2(a^{2}-4a)+1=2×(-1)+1=-1$.
请你根据小明的分析过程,解决下面的问题.
(1)化简:$\frac{2}{\sqrt{5}-\sqrt{3}}$; (2)若$a=\frac{1}{\sqrt{2}-1}$,求$3a^{2}-6a-1$的值.
$\because a=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3},\therefore a-2=-\sqrt{3}$,
$\therefore(a-2)^{2}=3$,即$a^{2}-4a+4=3,\therefore a^{2}-4a=-1$,
$\therefore2a^{2}-8a+1=2(a^{2}-4a)+1=2×(-1)+1=-1$.
请你根据小明的分析过程,解决下面的问题.
(1)化简:$\frac{2}{\sqrt{5}-\sqrt{3}}$; (2)若$a=\frac{1}{\sqrt{2}-1}$,求$3a^{2}-6a-1$的值.
答案：12.解:(1)$\frac{2}{\sqrt{5}-\sqrt{3}}=\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\sqrt{5}+\sqrt{3}$.
(2)$\because a=\frac{\sqrt{2}+1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}+1,\therefore a-1=\sqrt{2}$,
$\therefore (a-1)^2=2$,即$a^2-2a+1=2,\therefore a^2-2a=1$,
$\therefore 3a^2-6a-1=3(a^2-2a)-1=3×1-1=2$.
(2)$\because a=\frac{\sqrt{2}+1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}+1,\therefore a-1=\sqrt{2}$,
$\therefore (a-1)^2=2$,即$a^2-2a+1=2,\therefore a^2-2a=1$,
$\therefore 3a^2-6a-1=3(a^2-2a)-1=3×1-1=2$.