\(\left\{{}\begin{matrix}x+y+z=6\\x^2+y^2+z^2=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+4y+4z=24\\x^2+y^2+z^2=12\end{matrix}\right.\Leftrightarrow x^2-4x+y^2-4y+z^2-4z=-12\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+\left(z^2-4x+4\right)=0\Leftrightarrow\left(x-2\right)^2+\left(y-2\right)^2+\left(z-2\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-2\right)^2=0\\\left(z-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-2=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\\z=2\end{matrix}\right..Vậy:x=y=z=2\)
