Question

Cho a,b.c đôi một khác nhau, thỏa mãn a³ + b³ + c³ = 3abc và abc khác 0

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

Answer 1

\(a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\)
\(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+2ab=c^2\\a^2+c^2+2ac=b^2\\b^2+c^2+2cb=a^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2-c^2=-2ab\\a^2+c^2-b^2=-2ac\\b^2+c^2-a^2=-2cb\end{matrix}\right.\)
\(\Rightarrow P=\frac{ab^2}{-2ab}+\frac{bc^2}{-2bc}+\frac{ca^2}{-2ac}=\frac{-b}{2}+\frac{-c}{2}+\frac{-a}{2}=\frac{-\left(a+b+c\right)}{2}=\frac{0}{2}=0\)