Question

cho A=\(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)

rut gon A biet a+b+c=0

Answer 1

Từ \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=c^2\\a^2+2ac+c^2=b^2\\b^2+2bc+c^2=a^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2-c^2=-2ab\\a^2+c^2-c^2=-2ac\\b^2+c^2-a^2=-2bc\\\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{1}{-2ab}+\dfrac{1}{-2ac}+\dfrac{1}{-2bc}=\dfrac{a+b+c}{-2abc}=\dfrac{0}{-2abc}=0\)