Lời giải:
Do \(a+b+c=0\rightarrow b+c=-a\), suy ra:
\(b^2+c^2-a^2=(b+c)^2-a^2-2bc=(-a)^2-a^2-2bc=-2bc\)
\(\Rightarrow \frac{1}{b^2+c^2-a^2}=\frac{1}{-2bc}\)
Tương tự với các phân thức còn lại:
\(\Rightarrow B=\frac{1}{-2bc}+\frac{1}{-2ac}+\frac{1}{-2ab}=\frac{-1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=\frac{-1}{2}.\frac{a+b+c}{abc}=0\)
Vậy \(B=0\)
Từ \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow a^2+b^2+c^2=-2ab-2bc-2ca\)
Khi đó \(\dfrac{1}{b^2+c^2-a^2}=\dfrac{1}{-2ab-2bc-2ca-2a^2}=\dfrac{1}{-2\left(a+b\right)\left(a+c\right)}\)
Viết lại \(B=-\dfrac{1}{2}\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(b+c\right)\left(a+b\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(=-\dfrac{1}{2}\cdot\dfrac{b+c+c+a+a+b}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)\(=-\dfrac{1}{2}\cdot\dfrac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=0\)