\(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)
\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow2\left(x-1\right)^2+3\left(x+y\right)^2+\left(z+2\right)^2=0\)
Vì \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\3\left(x+y\right)^2\ge0\\\left(z+2\right)^2\ge0\end{matrix}\right.\)\(\forall x;y;z\) Nên \(2\left(x-1\right)^2+3\left(x+y\right)^2+\left(z+2\right)^2\ge0\forall x;y;z\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)^2=0\\3\left(x+y\right)^2=0\\\left(z+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)
Vậy \(x=1;y=-1;z=-2\)
