19. (本题 8 分)
先化简,再求值：已知$(\frac{x + 1}{x - 1} + \frac{1}{x^{2} - 2x + 1}) ÷ \frac{x}{x - 1}$,其中$- 1 ＜ x ＜ 3$,选择一个整数$x$的值代入求值.

2

化简过程：
$\begin{aligned}&(\frac{x + 1}{x - 1} + \frac{1}{x^2 - 2x + 1}) ÷ \frac{x}{x - 1}\\=&\left[\frac{x + 1}{x - 1} + \frac{1}{(x - 1)^2}\right] × \frac{x - 1}{x}\\=&\left[\frac{(x + 1)(x - 1)}{(x - 1)^2} + \frac{1}{(x - 1)^2}\right] × \frac{x - 1}{x}\\=&\frac{x^2 - 1 + 1}{(x - 1)^2} × \frac{x - 1}{x}\\=&\frac{x^2}{(x - 1)^2} × \frac{x - 1}{x}\\=&\frac{x}{x - 1}\end{aligned}$
求值：
$\because -1 ＜ x ＜ 3$，且$x$为整数，分式有意义需$x \neq 0,1$，$\therefore x = 2$。
当$x = 2$时，$\frac{x}{x - 1} = \frac{2}{2 - 1} = 2$。

20. (本题 8 分)
探索发现：$\frac{1}{1 × 2} = 1 - \frac{1}{2}$；$\frac{1}{2 × 3} = \frac{1}{2} - \frac{1}{3}$；$\frac{1}{3 × 4} = \frac{1}{3} - \frac{1}{4}$；…
根据你发现的规律,回答下列问题.
(1)$\frac{1}{4 × 5}$ = ,$\frac{1}{n × (n + 1)}$ = .
(2)利用你发现的规律计算：$\frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} + x + \frac{1}{n × (n + 1)}$.

(3)灵活利用规律解方程：$\frac{1}{x(x + 2)} + \frac{1}{(x + 2)(x + 4)} + … + \frac{1}{(x + 98)(x + 100)} = \frac{1}{x + 100}$.
$\begin{aligned}&\frac{1}{2}×(\frac{2}{x(x + 2)}+\frac{2}{(x + 2)(x + 4)}+\cdots+\frac{2}{(x + 98)(x + 100)})\\=&\frac{1}{2}×[(\frac{1}{x}-\frac{1}{x + 2})+(\frac{1}{x + 2}-\frac{1}{x + 4})+\cdots+(\frac{1}{x + 98}-\frac{1}{x + 100})]\\=&\frac{1}{2}×(\frac{1}{x}-\frac{1}{x + 100})\\=&\frac{1}{x + 100}\end{aligned}$
则$\frac{1}{2}×(\frac{1}{x}-\frac{1}{x + 100})=\frac{1}{x + 100}$
$\frac{1}{x}-\frac{1}{x + 100}=\frac{2}{x + 100}$
$\frac{1}{x}=\frac{3}{x + 100}$
$x + 100 = 3x$
$2x=100$
$x = 50$
经检验，$x = 50$是原方程的解。

(1)
$\frac{1}{4×5}=\frac{1}{4}-\frac{1}{5}$；
$\frac{1}{n(n + 1)}=\frac{1}{n}-\frac{1}{n + 1}$。
(2)
$\begin{aligned}&\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\cdots+\frac{1}{n(n + 1)}\\=&(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots+(\frac{1}{n}-\frac{1}{n + 1})\\=&1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{n}-\frac{1}{n + 1}\\=&1-\frac{1}{n + 1}\\=&\frac{n}{n + 1}\end{aligned}$
(3)
$\begin{aligned}&\frac{1}{2}×(\frac{2}{x(x + 2)}+\frac{2}{(x + 2)(x + 4)}+\cdots+\frac{2}{(x + 98)(x + 100)})\\=&\frac{1}{2}×[(\frac{1}{x}-\frac{1}{x + 2})+(\frac{1}{x + 2}-\frac{1}{x + 4})+\cdots+(\frac{1}{x + 98}-\frac{1}{x + 100})]\\=&\frac{1}{2}×(\frac{1}{x}-\frac{1}{x + 100})\\=&\frac{1}{x + 100}\end{aligned}$
则$\frac{1}{2}×(\frac{1}{x}-\frac{1}{x + 100})=\frac{1}{x + 100}$
$\frac{1}{x}-\frac{1}{x + 100}=\frac{2}{x + 100}$
$\frac{1}{x}=\frac{3}{x + 100}$
$x + 100 = 3x$
$2x=100$
$x = 50$
经检验，$x = 50$是原方程的解。
综上，答案依次为：(1)$\frac{1}{4}-\frac{1}{5}$；$\frac{1}{n}-\frac{1}{n + 1}$；(2)$\frac{n}{n + 1}$；(3)$x = 50$。