\(x^2+y^2+z^2+2x-4y+6z=-14\\ x^2+2x+1+y^2-4y+4+z^2+6z+9=0\\ \left(x+1\right)^2+\left(y-4\right)^2+\left(z+3\right)^2=0\\ \Rightarrow\left\{{}\begin{matrix}x+1=0\\y-4=0\\z+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=4\\z=-3\end{matrix}\right.\\ \Rightarrow x+y+z=-1+4-3=0\)
có \(x^2+y^2+z^2+2x-4y+6z=-14\)
=>\(x^2+z^2+y^2+2x-4y+6z+14=0\)
=>\(\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2+6y+9\right)=0\)
=>\(\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
=> x+1 =0
y-2 =0
z+3 =0
=> x = -1
y = 2
z = -3
=> x + y + z = -2
