\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Ta có : \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\\\left(z+1\right)^2\ge0\end{cases}\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0}\)
Do đó dấu đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\\left(z+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
Vậy nghiệm của phương trình là : \(\left(x;y;z\right)=\left(1;3;-1\right)\)
