Từ \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
Ta có: \(B=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
\(=\frac{ab}{a^2+\left(b-c\right)\left(b+c\right)}+\frac{bc}{b^2+\left(c-a\right)\left(c+a\right)}+\frac{ca}{c^2+\left(a-b\right)\left(a+b\right)}\)
\(=\frac{ab}{a^2-a\left(b-c\right)}+\frac{bc}{b^2-b\left(c-a\right)}+\frac{ca}{c^2-c\left(a-b\right)}\)
\(=\frac{ab}{a\left(a-b+c\right)}+\frac{bc}{b\left(b-c+a\right)}+\frac{ca}{c\left(c-a+b\right)}\)
\(=\frac{b}{a-b+c}+\frac{c}{b-c+a}+\frac{a}{c-a+b}\)
\(=\frac{b}{a+c-b}+\frac{c}{b+a-c}+\frac{a}{c+b-a}\)
\(=\frac{b}{-2b}+\frac{c}{-2c}+\frac{a}{-2a}=\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)=-\frac{3}{2}\)
