Question

Cho x, y, z là các số thưc thỏa mãn: \(2x^2+2y^2+z^2-2x+2y+2xy+2yz+2zx+2=0\)

Tìm giá trị biểu thức A= \(x^{2018}+y^{2018}+z^{2018}\)

Answer 1

\(x^2+y^2+z^2+2xy+2yz+2xz+x^2-2x+1+y^2+2y+1=0\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x-1=0\\y+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}z=-\left(x+y\right)\\x=1\\y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=0\end{matrix}\right.\)

\(\Rightarrow A=1^{2018}+\left(-1\right)^{2018}+0^{2018}=1+1+0=2\)