23. (本题满分 13 分)
(1)观察下列算式:$\frac {1}{6}=\frac {1}{2×3}=\frac {1}{2}-\frac {1}{3},\frac {1}{12}=\frac {1}{3×4}=\frac {1}{3}-\frac {1}{4},\frac {1}{20}=\frac {1}{4×5}=\frac {1}{4}-\frac {1}{5},... $由此可推断$\frac {1}{42}=$;
(2)请用含字母$m$($m$为正整数)的等式表示(1)中的一般规律;
(3)解方程:$\frac {1}{(x-2)(x-3)}-\frac {3}{(x-1)(x-4)}+\frac {1}{(x-1)(x-2)}=\frac {1}{x-4}$。

(1)
$\frac{1}{42}=\frac{1}{6×7}=\frac{1}{6}-\frac{1}{7}$
(2)
$\frac{1}{m(m + 1)}=\frac{1}{m}-\frac{1}{m + 1}$
(3)
方程$\frac{1}{(x - 2)(x - 3)}-\frac{3}{(x - 1)(x - 4)}+\frac{1}{(x - 1)(x - 2)}=\frac{1}{x - 4}$
可化为$(\frac{1}{x - 3}-\frac{1}{x - 2})-\ 3×\frac{1}{3}(\frac{1}{x - 4}-\frac{1}{x - 1})+\frac{1}{x - 2}-\frac{1}{x - 1}=\frac{1}{x - 4}$
$(\frac{1}{x - 3}-\frac{1}{x - 2})-(\frac{1}{x - 4}-\frac{1}{x - 1})+(\frac{1}{x - 2}-\frac{1}{x - 1})=\frac{1}{x - 4}$
$\frac{1}{x - 3}-\frac{1}{x - 4}=\frac{1}{x - 4}$
$\frac{1}{x - 3}=\frac{2}{x - 4}$
$x - 4 = 2(x - 3)$
$x - 4 = 2x-6$
$2x - x = 2$
$x = 2$
检验：当$x = 2$时，$(x - 2)(x - 3)(x - 1)(x - 4)=0$，所以$x = 2$是增根，原方程无解。