Tìm x , y , z thỏa mãn phương trình sau : 9x^2+y^2+2z^2-18x+4z-6y+20=0

$9x^{2}+$ $y^{2}+$ $2z^{2}-18x+4z-6y +20=0$

$⇔(x^{2}-18x+9)+($ $y^{2}-6x+9)+($ $2z^{2}+4z+2)=0$

$⇔9(^{}$ $x^{2}-2x+1)+($ $y^{2}-6y+9)+2($ $z^{2}+2z+1)=0$

$⇔9(^{}$ $x-1)^{2}+(y-3)$$^{3}+2(z+1)$$^{2}=0$

$mà\left[\begin{array}{ccc}9(x-1)^{2}\geq0\\(y-3)^{2}\geq0\\2(z+1)^{2}\geq0\end{array}\right]$

$\text{ ⇒ 9 ( x - 1 )}$$^{2}+(y-3)$$^{2}+2(z+1)$$^{2}$ $\geq0$

$dấu = xảy ra ⇔\left[\begin{array}{ccc}9(x-1)^{2}=0\\(y-3)^{2}=0\\2(z+1)^{2}=0\end{array}\right]$

$⇒\left[\begin{array}{ccc}x-1=0\\y-3=0\\z+1=0\end{array}\right]⇒$ $\left[\begin{array}{ccc}x=1\\y=3\\z=-1\end{array}\right]$

$\text{ vậy x = 1 , y = 3 , z = - 1 . }$