Question

Cho a+b+c=0 ( abc khác 0). Rút gon:

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

Answer 1

Từ \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

Đầu tiên ta có hẳng đẳng thức:

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

\(\Rightarrow0=a^3+b^3+c^3+3\left(-c\right)\left(-b\right)\left(-a\right)\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\Rightarrow a^3+b^3+c^3=3abc\)

Biến đổi mẫu thức:

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

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

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