Giải thích các bước giải:
`9x^2 + y^2 + 2z^2 - 18x + 4z - 6y + 20 = 0`
`=>(9x^2- 18x+9) + (y^2- 6y+9) + (2z^2 + 4z + 2) = 0`
`=>9(x^2- 2x+1) + (y^2- 6y+9) + 2(z^2 + 2z + 1) = 0`
`=>9(x-1)^2+(y-3)^2+2(z+1)^2=0`
Mà $\left\{\begin{matrix}9(x-1)^2\ge0\\(y-3)^2\ge0\\2(z+1)^2\ge0\end{matrix}\right.$
`=>9(x-1)^2+(y-3)^2+2(z+1)^2>=0`
Dấu `=` xảy ra `<=>`$\left\{\begin{matrix}9(x-1)^2=0\\(y-3)^2=0\\2(z+1)^2=0\end{matrix}\right.$
`=>`$\left\{\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.$
Vậy `x=1;y=3;z=-1.`
