21. 先化简，再求值：
(1) $\frac{x}{x - 4} · (\frac{x + 2}{x^{2} - 2x} - \frac{x - 1}{x^{2} - 4x + 4})$，其中$x = 3 - \sqrt{2}$；
(2) $\frac{x - 2}{x - 1} ÷ (x + 1 - \frac{3}{x - 1})$，其中$x = \sqrt{5} - 4$。
答案：21.(1)原式$ = \frac{1}{(x - 2)^2}$.当$x = 3 - \sqrt{2}$时，原式$ = \frac{1}{(3 - \sqrt{2} - 2)^2} = \frac{1}{(1 - \sqrt{2})^2} = \frac{1}{3 - 2\sqrt{2}} = 3 + 2\sqrt{2}$ (2)原式$ = \frac{1}{x + 2}$.当$x = \sqrt{5} - 4$时，原式$ = \frac{1}{\sqrt{5} - 4 + 2} = \frac{1}{\sqrt{5} - 2} = \sqrt{5} + 2$
解析：
解：原式$=\frac{x}{x - 4}·[\frac{x + 2}{x(x - 2)} - \frac{x - 1}{(x - 2)^2}]$
$=\frac{x}{x - 4}·[\frac{(x + 2)(x - 2) - x(x - 1)}{x(x - 2)^2}]$
$=\frac{x}{x - 4}·[\frac{x^2 - 4 - x^2 + x}{x(x - 2)^2}]$
$=\frac{x}{x - 4}·\frac{x - 4}{x(x - 2)^2}$
$=\frac{1}{(x - 2)^2}$
当$x = 3 - \sqrt{2}$时，原式$=\frac{1}{(3 - \sqrt{2} - 2)^2}=\frac{1}{(1 - \sqrt{2})^2}=\frac{1}{3 - 2\sqrt{2}}=3 + 2\sqrt{2}$
22. (新考法·阅读理解)阅读材料：
将边长分别为$a$，$a + \sqrt{b}$，$a + 2\sqrt{b}$，$a + 3\sqrt{b}$的正方形的面积分别记为$S_{1}$，$S_{2}$，$S_{3}$，$S_{4}$，则$S_{2} - S_{1} = (a + \sqrt{b})^{2} - a^{2} = [(a + \sqrt{b}) + a] · [(a + \sqrt{b}) - a] = (2a + \sqrt{b}) · \sqrt{b} = b + 2a\sqrt{b}$。例如：当$a = 1$，$b = 3$时，$S_{2} - S_{1} = 3 + 2\sqrt{3}$。
根据以上材料，解答下列问题：
(1) 当$a = 1$，$b = 3$时，$S_{3} - S_{2} =$
$9 + 2\sqrt{3}$
，$S_{4} - S_{3} =$
$15 + 2\sqrt{3}$
。
(2) 当$a = 1$，$b = 3$时，把边长为$a + n\sqrt{b}$的正方形的面积记作$S_{n + 1}$，其中$n$是正整数，请你根据(1)中的计算结果猜测$S_{n + 1} - S_{n}$等于多少，并证明你的猜想。
(3) 当$a = 1$，$b = 3$时，令$t_{1} = S_{2} - S_{1}$，$t_{2} = S_{3} - S_{2}$，$t_{3} = S_{4} - S_{3}$，$···$，$t_{n} = S_{n + 1} - S_{n}$，且$T = t_{1} + t_{2} + t_{3} + ··· + t_{50}$，求$T$的值。
答案：22.(1)$9 + 2\sqrt{3}$ $15 + 2\sqrt{3}$ (2)$S_{n + 1} - S_n = 6n - 3 + 2\sqrt{3}$ $S_{n + 1} - S_n = (1 + \sqrt{3}n)^2 - [1 + (n - 1)\sqrt{3}]^2 = [2 + (2n - 1) · \sqrt{3}] × \sqrt{3} = 3(2n - 1) + 2\sqrt{3} = 6n - 3 + 2\sqrt{3}$ (3)当$a = 1$,$b = 3$时，$T = t_1 + t_2 + t_3 + ··· + t_{50} = S_2 - S_1 + S_3 - S_2 + S_4 - S_3 + ··· + S_{51} - S_{50} = S_{51} - S_1 = (1 + 50\sqrt{3})^2 - 1 = 7500 + 100\sqrt{3}$
解析：
(1) $9 + 2\sqrt{3}$；$15 + 2\sqrt{3}$
(2) 猜想：$S_{n + 1} - S_n = 6n - 3 + 2\sqrt{3}$
证明：$S_{n + 1} - S_n = (1 + n\sqrt{3})^2 - [1 + (n - 1)\sqrt{3}]^2$
$=[(1 + n\sqrt{3}) + (1 + (n - 1)\sqrt{3})][(1 + n\sqrt{3}) - (1 + (n - 1)\sqrt{3})]$
$=[2 + (2n - 1)\sqrt{3}] · \sqrt{3}$
$=2\sqrt{3} + 3(2n - 1)$
$=6n - 3 + 2\sqrt{3}$
(3) $T = t_1 + t_2 + ··· + t_{50} = (S_2 - S_1) + (S_3 - S_2) + ··· + (S_{51} - S_{50}) = S_{51} - S_1$
$=(1 + 50\sqrt{3})^2 - 1^2 = 1 + 100\sqrt{3} + 7500 - 1 = 7500 + 100\sqrt{3}$
