Ta có: \(a+b+c=0\)
\(\Rightarrow c=-\left(a+b\right)\)
\(\Rightarrow b=-\left(a+c\right)\)
\(\Rightarrow a=-\left(b+c\right)\)
Thay: \(\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\) vào \(M\) ta được:
\(M=\frac{1}{a^2+b^2-\left(a+b\right)^2}+\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{c^2+a^2-\left(a+c\right)^2}\)
\(=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}\)
\(=\frac{a+b+c}{-2abc}=0\)