20. 先阅读下面的解答过程，然后作答：
形如$\sqrt{m\pm2\sqrt{n}}$的化简，只要我们找到两个数$a$，$b$使$a + b = m$，$ab = n$，这样$(\sqrt{a})^{2} + (\sqrt{b})^{2} = m$，$\sqrt{a}·\sqrt{b} = \sqrt{n}$，那么便有$\sqrt{m\pm2\sqrt{n}} = \sqrt{(\sqrt{a}\pm\sqrt{b})^{2}} = \sqrt{a}\pm\sqrt{b}(a ＞ b)$.例如：化简$\sqrt{7 + 4\sqrt{3}}$.
解：首先把$\sqrt{7 + 4\sqrt{3}}$化为$\sqrt{7 + 2\sqrt{12}}$，这里$m = 7$，$n = 12$.$\because4 + 3 = 7$，$4×3 = 12$，即$(\sqrt{4})^{2} + (\sqrt{3})^{2} = 7$，$\sqrt{4}×\sqrt{3} = \sqrt{12}$，$\therefore\sqrt{7 + 4\sqrt{3}} = \sqrt{7 + 2\sqrt{12}} = \sqrt{(\sqrt{4} + \sqrt{3})^{2}} = 2 + \sqrt{3}$.
由上述例题的方法化简：
(1)$\sqrt{13 - 2\sqrt{42}}$；
(2)$\sqrt{7 - \sqrt{40}}$；
(3)$\sqrt{2 - \sqrt{3}}$.
形如$\sqrt{m\pm2\sqrt{n}}$的化简，只要我们找到两个数$a$，$b$使$a + b = m$，$ab = n$，这样$(\sqrt{a})^{2} + (\sqrt{b})^{2} = m$，$\sqrt{a}·\sqrt{b} = \sqrt{n}$，那么便有$\sqrt{m\pm2\sqrt{n}} = \sqrt{(\sqrt{a}\pm\sqrt{b})^{2}} = \sqrt{a}\pm\sqrt{b}(a ＞ b)$.例如：化简$\sqrt{7 + 4\sqrt{3}}$.
解：首先把$\sqrt{7 + 4\sqrt{3}}$化为$\sqrt{7 + 2\sqrt{12}}$，这里$m = 7$，$n = 12$.$\because4 + 3 = 7$，$4×3 = 12$，即$(\sqrt{4})^{2} + (\sqrt{3})^{2} = 7$，$\sqrt{4}×\sqrt{3} = \sqrt{12}$，$\therefore\sqrt{7 + 4\sqrt{3}} = \sqrt{7 + 2\sqrt{12}} = \sqrt{(\sqrt{4} + \sqrt{3})^{2}} = 2 + \sqrt{3}$.
由上述例题的方法化简：
(1)$\sqrt{13 - 2\sqrt{42}}$；
(2)$\sqrt{7 - \sqrt{40}}$；
(3)$\sqrt{2 - \sqrt{3}}$.
答案：$20.(1)\sqrt{13 - 2\sqrt{42}} = \sqrt{(\sqrt{7} - \sqrt{6})^2} = \sqrt{7} - \sqrt{6} (2)\sqrt{7 - \sqrt{40}} =$
$\sqrt{7 - 2\sqrt{10}} = \sqrt{(\sqrt{5} - \sqrt{2})^2} = \sqrt{5} - \sqrt{2} (3)\sqrt{2 - \sqrt{3}} = \sqrt{\frac{8 - 4\sqrt{3}}{4}} =$
$\sqrt{\frac{8 - 2\sqrt{12}}{4}} = \sqrt{\frac{(\sqrt{6} - \sqrt{2})^2}{4}} = \frac{\sqrt{6} - \sqrt{2}}{2}$
$\sqrt{7 - 2\sqrt{10}} = \sqrt{(\sqrt{5} - \sqrt{2})^2} = \sqrt{5} - \sqrt{2} (3)\sqrt{2 - \sqrt{3}} = \sqrt{\frac{8 - 4\sqrt{3}}{4}} =$
$\sqrt{\frac{8 - 2\sqrt{12}}{4}} = \sqrt{\frac{(\sqrt{6} - \sqrt{2})^2}{4}} = \frac{\sqrt{6} - \sqrt{2}}{2}$
21. 解决问题：若$a = \dfrac{1}{2 + \sqrt{3}}$，求$2a^{2} - 8a + 1$的值.小娟是这样分析与解答的：
$\because a = \dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})×(2 - \sqrt{3})} = 2 - \sqrt{3}$，$\therefore a - 2 = -\sqrt{3}$.$\therefore(a - 2)^{2} = 3$.$\therefore a^{2} - 4a + 4 = 3$.$\therefore a^{2} - 4a = - 1$.$\therefore2a^{2} - 8a + 1 = 2(a^{2} - 4a) + 1 = 2×(-1) + 1 = - 1$.
请根据小娟的分析解答过程，解决以下问题：
(1)计算：$\dfrac{1}{\sqrt{3} + 1} + \dfrac{1}{\sqrt{5} + \sqrt{3}} + \dfrac{1}{\sqrt{7} + \sqrt{5}} + ··· + \dfrac{1}{\sqrt{121} + \sqrt{119}}$.
(2)已知$a = \dfrac{1}{\sqrt{2} - 1}$.
①求$3a^{2} - 6a + 1$的值.
②直接写出代数式的值：$a^{3} - 3a^{2} + a + 1 =$
$\because a = \dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})×(2 - \sqrt{3})} = 2 - \sqrt{3}$，$\therefore a - 2 = -\sqrt{3}$.$\therefore(a - 2)^{2} = 3$.$\therefore a^{2} - 4a + 4 = 3$.$\therefore a^{2} - 4a = - 1$.$\therefore2a^{2} - 8a + 1 = 2(a^{2} - 4a) + 1 = 2×(-1) + 1 = - 1$.
请根据小娟的分析解答过程，解决以下问题：
(1)计算：$\dfrac{1}{\sqrt{3} + 1} + \dfrac{1}{\sqrt{5} + \sqrt{3}} + \dfrac{1}{\sqrt{7} + \sqrt{5}} + ··· + \dfrac{1}{\sqrt{121} + \sqrt{119}}$.
(2)已知$a = \dfrac{1}{\sqrt{2} - 1}$.
①求$3a^{2} - 6a + 1$的值.
②直接写出代数式的值：$a^{3} - 3a^{2} + a + 1 =$
0
；$2a^{2} - 5a + \dfrac{1}{a} + 2 =$2
.答案：21.(1)原式$=\frac{\sqrt{3} - 1}{(\sqrt{3} + 1)(\sqrt{3} - 1)} + \frac{\sqrt{5} - \sqrt{3}}{(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})} +$
$\frac{\sqrt{7} - \sqrt{5}}{(\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5})} + ··· + \frac{\sqrt{121} - \sqrt{119}}{(\sqrt{121} + \sqrt{119})(\sqrt{121} - \sqrt{119})} =$
$\frac{\sqrt{3} - 1}{2} + \frac{\sqrt{5} - \sqrt{3}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2} + ··· + \frac{\sqrt{121} - \sqrt{119}}{2} = \frac{1}{2}×(\sqrt{3} -$
$1 + \sqrt{5} - \sqrt{3} + \sqrt{7} - \sqrt{5} + ··· + \sqrt{121} - \sqrt{119}) = \frac{1}{2}×(\sqrt{121} -$
1)=5 (2)①
∵$a = \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \sqrt{2} + 1，$
∴$a - 1 = \sqrt{2}.$
∴$(a - 1)^2 = 2.$
∴$a^2 - 2a + 1 = 2.$
∴$a^2 - 2a = 1.$
∴$3a^2 - 6a + 1 = 3(a^2 - 2a) + 1 = 3×1 + 1 = 4 ②0 2$
$\frac{\sqrt{7} - \sqrt{5}}{(\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5})} + ··· + \frac{\sqrt{121} - \sqrt{119}}{(\sqrt{121} + \sqrt{119})(\sqrt{121} - \sqrt{119})} =$
$\frac{\sqrt{3} - 1}{2} + \frac{\sqrt{5} - \sqrt{3}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2} + ··· + \frac{\sqrt{121} - \sqrt{119}}{2} = \frac{1}{2}×(\sqrt{3} -$
$1 + \sqrt{5} - \sqrt{3} + \sqrt{7} - \sqrt{5} + ··· + \sqrt{121} - \sqrt{119}) = \frac{1}{2}×(\sqrt{121} -$
1)=5 (2)①
∵$a = \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \sqrt{2} + 1，$
∴$a - 1 = \sqrt{2}.$
∴$(a - 1)^2 = 2.$
∴$a^2 - 2a + 1 = 2.$
∴$a^2 - 2a = 1.$
∴$3a^2 - 6a + 1 = 3(a^2 - 2a) + 1 = 3×1 + 1 = 4 ②0 2$