Do \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}b+c=-a\\a+c=-b\end{matrix}\right.\)
\(a^2+b^2-c^2=a^2+\left(b+c\right)\left(b-c\right)=a^2-a\left(b-c\right)=a\left(a+c-b\right)=a\left(-2b\right)=-2ab\)
Tương tự ta có:
\(b^2+c^2-a^2=-2bc\)
\(c^2+a^2-b^2=-2ac\)
\(\Rightarrow A=\dfrac{1}{-2ab}+\dfrac{1}{-2bc}+\dfrac{1}{-2ac}=\dfrac{a+b+c}{-2abc}=0\)