1. 定义：我们将$(\sqrt{a}+\sqrt{b})$与$(\sqrt{a}-\sqrt{b})$称为一对“对偶式”. 因为$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = (\sqrt{a})^{2}-(\sqrt{b})^{2}=a - b$，可以有效去掉根号，所以有一些问题可以通过构造“对偶式”来解决.
例如：已知$\sqrt{18 - x}-\sqrt{11 - x}=1$，求$\sqrt{18 - x}+\sqrt{11 - x}$的值，可以这样解答：
因为$(\sqrt{18 - x}-\sqrt{11 - x})\times(\sqrt{18 - x}+\sqrt{11 - x})=(\sqrt{18 - x})^{2}-(\sqrt{11 - x})^{2}=18 - x - 11 + x = 7$，
所以$\sqrt{18 - x}+\sqrt{11 - x}=7$.
(1)已知：$\sqrt{20 - x}+\sqrt{4 - x}=8$，求$\sqrt{20 - x}-\sqrt{4 - x}$的值；
(2)结合已知条件和第(1)问的结果，解方程：$\sqrt{20 - x}+\sqrt{4 - x}=8$；
(3)计算：$\frac{1}{3\sqrt{1}+\sqrt{3}}+\frac{1}{5\sqrt{3}+3\sqrt{5}}+\frac{1}{7\sqrt{5}+5\sqrt{7}}+\cdots+\frac{1}{2023\sqrt{2021}+2021\sqrt{2023}}$.

(1) $\because (\sqrt{20 - x} + \sqrt{4 - x})(\sqrt{20 - x} - \sqrt{4 - x}) = (\sqrt{20 - x})^2 - (\sqrt{4 - x})^2 = 20 - x - 4 + x = 16$，且$\sqrt{20 - x} + \sqrt{4 - x} = 8$，$\therefore \sqrt{20 - x} - \sqrt{4 - x} = 2$.
(2) $\because \begin{cases}\sqrt{20 - x} + \sqrt{4 - x} = 8\\\sqrt{20 - x} - \sqrt{4 - x} = 2\end{cases}$，$\therefore 2\sqrt{4 - x} = 6$，化简后两边同时平方得$4 - x = 9$，$\therefore x = - 5$，经检验：$x = - 5$是原方程的解.
(3) $\frac{1}{3\sqrt{1}+\sqrt{3}} + \frac{1}{5\sqrt{3}+3\sqrt{5}} + \frac{1}{7\sqrt{5}+5\sqrt{7}} + \cdots + \frac{1}{2023\sqrt{2021}+2021\sqrt{2023}} = \frac{3 - \sqrt{3}}{6} + \frac{5\sqrt{3}-3\sqrt{5}}{30} + \frac{7\sqrt{5}-5\sqrt{7}}{70} + \cdots + \frac{2023\sqrt{2021}-2021\sqrt{2023}}{2023\times2021 - 2021\times2023} = \frac{1}{2} - \frac{\sqrt{3}}{6} + \frac{\sqrt{3}}{6} - \frac{\sqrt{5}}{10} + \frac{\sqrt{5}}{10} - \frac{\sqrt{7}}{14} + \cdots + \frac{\sqrt{2021}}{4042} - \frac{\sqrt{2023}}{4046} = \frac{1}{2} - \frac{\sqrt{2023}}{4046}$.

2. (2024·南通期末)阅读：如果两个分式$A$与$B$的和为常数$k$，且$k$为正整数，那么称$A$与$B$互为“关联分式”，常数$k$称为“关联值”. 如分式$A=\frac{x}{x - 1}$，$B=\frac{-1}{x - 1}$，$A + B=\frac{x - 1}{x - 1}=1$，则$A$与$B$互为“关联分式”，“关联值”$k = 1$.
(1)若分式$A=\frac{x - 4}{x - 3}$，$B=\frac{x - 2}{x - 3}$，判断$A$与$B$是否互为“关联分式”，若不是，请说明理由；若是，请求出“关联值”$k$.
(2)已知分式$C=\frac{2x - 1}{x - 3}$，$D=\frac{M}{x^{2}-9}$，$C$与$D$互为“关联分式”，且“关联值”$k = 2$.
①$M =$_______(用含$x$的式子表示)；
②若$x$为正整数，且分式$D$的值为正整数，则$x$的值等于_______.
(3)若分式$E=\frac{(x - a)(x - b)}{x - 4}$，$F=\frac{(x - c)(x - 5)}{4 - x}$（$a$、$b$为整数且$c = a + b$），$E$是$F$的“关联分式”，且“关联值”$k = 5$，求$c$的值.

(1) $A$与$B$是互为“关联分式”，理由如下：$\because A = \frac{x - 4}{x - 3}$，$B = \frac{x - 2}{x - 3}$，$\therefore A + B = \frac{x - 4}{x - 3} + \frac{x - 2}{x - 3} = \frac{2x - 6}{x - 3} = 2$，$\therefore A$与$B$是互为“关联分式”，“关联值”$k = 2$.
(2) ①$-15 - 5x$ 解析：$\because C = \frac{2x - 1}{x - 3}$，$D = \frac{M}{x^2 - 9}$，$\therefore C + D = \frac{(2x - 1)(x + 3)}{(x - 3)(x + 3)} + \frac{M}{(x - 3)(x + 3)} = \frac{2x^2 + 5x - 3 + M}{(x - 3)(x + 3)}$.$\because C$与$D$互为“关联分式”，且“关联值”$k = 2$，$\therefore 2x^2 + 5x - 3 + M = 2(x - 3)(x + 3) = 2x^2 - 18$，$\therefore M = 2x^2 - 18 - (2x^2 + 5x - 3) = - 15 - 5x$.
②2 解析：$\because D = \frac{M}{x^2 - 9} = \frac{-5(x + 3)}{(x + 3)(x - 3)} = -\frac{5}{x - 3}$，且分式$D$的值为正整数，$x$为正整数，$\therefore x - 3 = - 1$或$x - 3 = - 5$，$\therefore x = 2(x = - 2舍去)$.
(3) $E + F = \frac{(x - a)(x - b)}{x - 4} + \frac{(x - c)(x - 5)}{4 - x} = \frac{(x - a)(x - b)}{x - 4} - \frac{(x - c)(x - 5)}{x - 4} = \frac{(x - a)(x - b)-(x - c)(x - 5)}{x - 4} = \frac{-(a + b)x+ab+(5 + c)x - 5c}{x - 4}$.$\because c = a + b$，$k = 5$，$\therefore$原式$=\frac{-cx+ab+(5 + c)x - 5c}{x - 4} = \frac{5x+ab - 5c}{x - 4} = 5$，$\therefore ab - 5c = - 20$，即$ab - 5(a + b) = - 20$，$\therefore a(b - 5) = 5b - 20$，$\therefore a = \frac{5b - 25 + 5}{b - 5} = 5+\frac{5}{b - 5}$.$\because a$，$b$为整数，$\therefore b - 5$一定为$5$的约数，$\therefore b - 5 = - 1$或$- 5$或$1$或$5$，解得$b = 4$或$0$或$6$或$10$，$\therefore a = 0$或$4$或$10$或$6$，$\therefore c = a + b = 4 + 0 = 4$或$10 + 6 = 16$，$\therefore c$的值为$4$或$16$.