Ta có : \(A=x^2+y^2+z^2-2x-4y+6z=-14\)
\(\Leftrightarrow x^2+y^2+z^2+2x-4y+6z+14=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2+6z+9\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\left(1\right)\)
Do \(\left(x+1\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y;\left(z+3\right)^2\ge0\forall z\)
\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2\ge0\forall x;y;z\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-2=0\\z+3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\\z=-3\end{matrix}\right.\)
\(\Rightarrow x+y+z=-1+2-3=-2\)
Vậy \(x+y+z=-2\)
