Ta có: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2=c^2\\\left(a+c\right)^2=b^2\\\left(b+c\right)^2=a^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2-c^2=-2ab\\a^2+c^2-b^2=-2ac\\b^2+c^2-a^2=-2bc\end{matrix}\right.\)
Ta có: \(P=\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+b^2-c^2}+\frac{1}{a^2+c^2-b^2}\)
\(=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}\)
\(=-\left(\frac{1}{2ab}+\frac{1}{2bc}+\frac{1}{2ac}\right)\)
\(=\frac{-\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}{2}\)
\(=-\left[\left(\frac{1}{ab}+\frac{1}{ac}\right)+\left(\frac{1}{ab}+\frac{1}{bc}\right)+\left(\frac{1}{bc}+\frac{1}{ac}\right)\right]\)
\(=-\left[\frac{c+b}{abc}+\frac{c+a}{abc}+\frac{a+b}{abc}\right]\)
\(=-\frac{2\left(a+b+c\right)}{abc}\)
\(=-\frac{2\cdot0}{abc}=0\)
