20. 计算：
(1)$(\sqrt{12} + \sqrt{8})×\sqrt{3} - \sqrt{6}$；
(2)$\sqrt{27}÷3 - \sqrt{\dfrac{3}{2}}×\sqrt{12} + \sqrt{72}$；
(3)$(\sqrt{2} - 1)^{2} + \sqrt{32} - (\sqrt{5} + 3)(\sqrt{5} - 3)$；
(4)$(\sqrt{3} + 1)(\sqrt{3} - 1) - (\sqrt{18} - \sqrt{24})÷\sqrt{6}$.
(1)$(\sqrt{12} + \sqrt{8})×\sqrt{3} - \sqrt{6}$；
(2)$\sqrt{27}÷3 - \sqrt{\dfrac{3}{2}}×\sqrt{12} + \sqrt{72}$；
(3)$(\sqrt{2} - 1)^{2} + \sqrt{32} - (\sqrt{5} + 3)(\sqrt{5} - 3)$；
(4)$(\sqrt{3} + 1)(\sqrt{3} - 1) - (\sqrt{18} - \sqrt{24})÷\sqrt{6}$.
答案：20. (1)$6+\sqrt{6}$ (2)$\sqrt{3}+3\sqrt{2}$ (3)$7+2\sqrt{2}$ (4)$4-\sqrt{3}$
解析：
(1)
$\begin{aligned}&(\sqrt{12} + \sqrt{8})×\sqrt{3} - \sqrt{6}\\=&\sqrt{12}×\sqrt{3} + \sqrt{8}×\sqrt{3} - \sqrt{6}\\=&\sqrt{36} + \sqrt{24} - \sqrt{6}\\=&6 + 2\sqrt{6} - \sqrt{6}\\=&6 + \sqrt{6}\end{aligned}$
(2)
$\begin{aligned}&\sqrt{27}÷3 - \sqrt{\dfrac{3}{2}}×\sqrt{12} + \sqrt{72}\\=&3\sqrt{3}÷3 - \sqrt{\dfrac{3}{2}×12} + 6\sqrt{2}\\=&\sqrt{3} - \sqrt{18} + 6\sqrt{2}\\=&\sqrt{3} - 3\sqrt{2} + 6\sqrt{2}\\=&\sqrt{3} + 3\sqrt{2}\end{aligned}$
(3)
$\begin{aligned}&(\sqrt{2} - 1)^{2} + \sqrt{32} - (\sqrt{5} + 3)(\sqrt{5} - 3)\\=&(2 - 2\sqrt{2} + 1) + 4\sqrt{2} - (5 - 9)\\=&3 - 2\sqrt{2} + 4\sqrt{2} - (-4)\\=&3 + 2\sqrt{2} + 4\\=&7 + 2\sqrt{2}\end{aligned}$
(4)
$\begin{aligned}&(\sqrt{3} + 1)(\sqrt{3} - 1) - (\sqrt{18} - \sqrt{24})÷\sqrt{6}\\=&(3 - 1) - (\sqrt{18}÷\sqrt{6} - \sqrt{24}÷\sqrt{6})\\=&2 - (\sqrt{3} - 2)\\=&2 - \sqrt{3} + 2\\=&4 - \sqrt{3}\end{aligned}$
$\begin{aligned}&(\sqrt{12} + \sqrt{8})×\sqrt{3} - \sqrt{6}\\=&\sqrt{12}×\sqrt{3} + \sqrt{8}×\sqrt{3} - \sqrt{6}\\=&\sqrt{36} + \sqrt{24} - \sqrt{6}\\=&6 + 2\sqrt{6} - \sqrt{6}\\=&6 + \sqrt{6}\end{aligned}$
(2)
$\begin{aligned}&\sqrt{27}÷3 - \sqrt{\dfrac{3}{2}}×\sqrt{12} + \sqrt{72}\\=&3\sqrt{3}÷3 - \sqrt{\dfrac{3}{2}×12} + 6\sqrt{2}\\=&\sqrt{3} - \sqrt{18} + 6\sqrt{2}\\=&\sqrt{3} - 3\sqrt{2} + 6\sqrt{2}\\=&\sqrt{3} + 3\sqrt{2}\end{aligned}$
(3)
$\begin{aligned}&(\sqrt{2} - 1)^{2} + \sqrt{32} - (\sqrt{5} + 3)(\sqrt{5} - 3)\\=&(2 - 2\sqrt{2} + 1) + 4\sqrt{2} - (5 - 9)\\=&3 - 2\sqrt{2} + 4\sqrt{2} - (-4)\\=&3 + 2\sqrt{2} + 4\\=&7 + 2\sqrt{2}\end{aligned}$
(4)
$\begin{aligned}&(\sqrt{3} + 1)(\sqrt{3} - 1) - (\sqrt{18} - \sqrt{24})÷\sqrt{6}\\=&(3 - 1) - (\sqrt{18}÷\sqrt{6} - \sqrt{24}÷\sqrt{6})\\=&2 - (\sqrt{3} - 2)\\=&2 - \sqrt{3} + 2\\=&4 - \sqrt{3}\end{aligned}$
21. 已知$x = 1 - \sqrt{2}$，$y = 1 + \sqrt{2}$，求：
(1)$\dfrac{1}{x} + \dfrac{1}{y}$的值；
(2)$x^{2} + y^{2} + xy + 2x - 2y$的值.
(1)$\dfrac{1}{x} + \dfrac{1}{y}$的值；
(2)$x^{2} + y^{2} + xy + 2x - 2y$的值.
答案：21. (1)$\because x = 1-\sqrt{2},y = 1+\sqrt{2},\therefore x + y = 2,x - y=-2\sqrt{2}$,
$xy=-1.\therefore\frac{1}{x}+\frac{1}{y}=\frac{x + y}{xy}=\frac{2}{-1}=-2$ (2)$x^{2}+y^{2}+xy + 2x-2y=(x + y)^{2}-xy+2(x - y)=2^{2}-(-1)+2×(-2\sqrt{2})=5-4\sqrt{2}$
$xy=-1.\therefore\frac{1}{x}+\frac{1}{y}=\frac{x + y}{xy}=\frac{2}{-1}=-2$ (2)$x^{2}+y^{2}+xy + 2x-2y=(x + y)^{2}-xy+2(x - y)=2^{2}-(-1)+2×(-2\sqrt{2})=5-4\sqrt{2}$
22. 写作业时，小明被一道题难住了：若$a = \dfrac{1}{3 + \sqrt{10}}$，求$a^{2} + 6a - 27$的值.
老师给予了必要的方法提示：不宜直接代入计算，需要先化简已知式，如$a = \dfrac{1}{2 + \sqrt{3}}$.
$\because a = \dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})×(2 - \sqrt{3})} = 2 - \sqrt{3}$，
$\therefore a - 2 = -\sqrt{3}$.
$···$
请你根据老师的提示解决下面的问题.
(1) 计算：$\dfrac{1}{3 + \sqrt{6}} =$
(2) 若$a = \dfrac{1}{3 + \sqrt{10}}$，求$a^{2} + 6a - 27$的值.
老师给予了必要的方法提示：不宜直接代入计算，需要先化简已知式，如$a = \dfrac{1}{2 + \sqrt{3}}$.
$\because a = \dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})×(2 - \sqrt{3})} = 2 - \sqrt{3}$，
$\therefore a - 2 = -\sqrt{3}$.
$···$
请你根据老师的提示解决下面的问题.
(1) 计算：$\dfrac{1}{3 + \sqrt{6}} =$
$\frac{3-\sqrt{6}}{3}$
；(2) 若$a = \dfrac{1}{3 + \sqrt{10}}$，求$a^{2} + 6a - 27$的值.
答案：22. (1)$\frac{3-\sqrt{6}}{3}$ (2)$\because a=\frac{1}{3+\sqrt{10}}=\frac{3-\sqrt{10}}{(3+\sqrt{10})×(3-\sqrt{10})}=\sqrt{10}-3,\therefore a + 3=\sqrt{10}.\therefore a^{2}+6a-27=(a + 3)^{2}-36=(\sqrt{10})^{2}-36=-26$