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

<=>9x²-18x+9+y²-6y+9+2z²+4z+2=0

<=>(3x-3)²+(y-3)²+2.(z²+2z+1)=0

<=>(3x-3)²+(y-3)²+2.(z+1)²=0

<=>3x-3=0 và y-3=0 và z+1=0

<=>x=1 và y=3 và z=-1

(9x²-18x+9)+(y²-6y+9)+2(z²+2z+1)=0\(\Rightarrow\)(3x-3)²+(y-3)²+2(z+1)²=0\(\Rightarrow\hept{\begin{cases}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)

ta co 9(x^2-2x+1) +( y^2 -6y +9) + 2(z^2 + 2z +1) = 0

suy ra 9(x-1)^2 + (y - 3 )^2 + 3( z-1)^2 = 0

suy ra x-1=0 ; y-3 =0 ; z-1=0

suy ra x=1;y=3; z=1

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

\(\Leftrightarrow\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=1\\y=3\\z=-1\end{cases}\)

9x² + y² + 2z² - 18x + 4z - 6y + 20 = 0

( 9x² -18x + 9) +( y² - 6y + 9) +2(z²+2z +1) = 0

( 3x-3)² + ( y-3)² + 2( z+1)² = 0

vì ( 3x-3)^2 , (y-3)^2 , 2( z+1)^2 >0 \(\Rightarrow\left(3x-3\right)^2=\left(y-3\right)^2=2\left(z+1\right)^2\))^2

\(\Leftrightarrow\hept{\begin{cases}3x-3=0\\y-3=0\\2\left(z+1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Rightarrow\left[\left(3x\right)^2-2.3x.3+9\right]+\left(y^2-2.y.3+9\right)+\left(2z^2+4z+2\right)=0\)

\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)

\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

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

\(9x^2+y^2+2z^2-18x+4z-6z+20=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)