\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Rightarrow\left[\left(3x\right)^2-2.3x.3+9\right]+\left(y^2-2.y.3+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Vì \(\left(3x-3\right)^2\ge0\) với mọi x
\(\left(y-3\right)^2\ge0\) với mọi y
\(2\left(z+1\right)^2\ge0\) với mọi z
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\) với mọi x, y, z
Mà \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-3=0\\y-3=0\\\left(z+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3\left(x-1\right)=0\\y=3\\z+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y=3\\z=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
Vậy x = 1 ; y = 3 ; z = -1
