Dạng 2: Đơn giản biểu thức.

Ta có: 

$\begin{array}{l}\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-c)(b-a)}\\ =\frac{b(b-c)-a(a-c)}{ab(a-b)(b-c)(a-c)}\\ =\frac{b^2-bc-a^2+ac}{ab(a-b)(b-c)(a-c)}\\ =\frac{(a-b)(-a-b+c)}{ab(a-b)(b-c)(a-c)}\\ =\frac{-a-b+c}{ab(b-c)(a-c)}(a\ne b)\end{array}$

Vậy:

$\begin{array}{l}\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-c)(b-a)}+\frac{1}{c(c-a)(c-b)}\\ =\frac{-a-b+c}{ab(b-c)(a-c)}+\frac{1}{c(c-a)(c-b)}\\ =\frac{-ac-bc+c^2+ab}{abc(a-c)(b-c)}\\ =\frac{(a-c)(b-c)}{abc(a-c)(b-c)}\\ =\frac{1}{abc}(a\ne c,b\ne c)\end{array}$

Do $a+b+c=0=>a+b=-c=>a^2+2ab+b^2=c^2$

Nên $\left\{\begin{array}{l}{a}^2+b^2-c^2=-2ab\\ b^2+c^2-a^2=-2bc\\ a^2+c^2-b^2=-2ac\end{array}\right.$

Vậy $\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}$

$\begin{array}{l}=-\left(\frac{1}{2ab}+\frac{1}{2bc}+\frac{1}{2ca}\right)\\ =-\frac{a+b+c}{2abc}\end{array}$