Question

Cho a, b, c \(\ne0\) và a+b+c=0. Tính :

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

Answer 1

Ta có: \(a+b+c=0\Rightarrow a^2=\left(b+c\right)^2\Rightarrow a^2-2bc=b^2+c^2\)

\(\Rightarrow a^2-b^2-c^2=a^2-a^2+2bc=2bc\)

Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

\(A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}\)

Lại có: \(a+b+c=0\Rightarrow-a=b+c\)

                                   \(\Rightarrow-a^3=b^3+c^3+3bc\left(b+c\right)\)

                                  \(\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)=3abc\left(b+c=-a\right)\)

=> \(A=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)