Question

\(Cho\)\(a+b+c=0;abc\ne0\) 

Tính giá trị biểu thức:

P = \(\frac{c^2}{a^2+b^2-c^2}+\frac{a^2}{b^2+c^2-a^2}+\frac{b^2}{c^2+a^2-b^2}\)

Answer 1

abc khác 0 nhé ạ

Answer 2

Do \(a+b+c=0\)

\(\Rightarrow c=-a-b\)

\(\Rightarrow c^2=a^2+2ab+b^2\)

Tương tự,ta có:

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

\(b^2=a^2+2ac+c^2\)

Thay vào bài toán,ta được:

\(P=\frac{c^2}{a^2+b^2-\left(a^2+2ab+b^2\right)}+\frac{a^2}{b^2+c^2-\left(b^2+2bc+c^2\right)}+\frac{b^2}{c^2+a^2-\left(a^2+2ac+c^2\right)}\)

\(P=\frac{-c^2}{2ab}+\frac{-a^2}{2bc}+\frac{-b^2}{2ac}\)

\(P=\frac{-\left(a^3+b^3+c^3\right)}{2abc}\)

Do \(a+b+c=0\Rightarrow-a=b+c\)

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

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

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

Khi đó,ta có:
\(P=\frac{-\left(3abc\right)}{2abc}=-\frac{3}{2}\)