Lời giải:
$\frac{1}{(a-b)(a-c)}+\frac{1}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)}=\frac{-1}{(a-b)(c-a)}+\frac{-1}{(a-b)(b-c)}+\frac{-1}{(c-a)(b-c)}$
$=\frac{-(b-c)}{(a-b)(b-c)(c-a)}+\frac{-(c-a)}{(a-b)(b-c)(c-a)}+\frac{-(a-b)}{(a-b)(b-c)(c-a)}$
$=\frac{-(b-c+c-a+a-b)}{(a-b)(b-c)(c-a)}=0$