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\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=1\\y=3\\z=-1\end{cases}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)(*)
Vì \(\left(x-1\right)\ge0;\left(y-3\right)^2\ge0;\left(z+1\right)^2\ge0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\y=3\\z=-1\end{cases}}}\)
pt ⇔ ( 9x2 - 18x + 9 ) + ( y2 - 6y + 9 ) + ( 2z2 + 4z + 2 ) = 0
⇔ 9( x2 - 2x + 1 ) + ( y - 3 )2 + 2( z2 + 2z + 1 ) = 0
⇔ 9( x - 1 )2 + ( y - 3 )2 + 2( z + 1 )2 = 0
Vì \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\\2\left(z+1\right)^2\ge0\forall z\end{cases}}\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\forall x,y,z\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
Vậy
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\Leftrightarrow\hept{\begin{cases}3x-3=0\\y-3=0\\2\left(z+1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(9x2-18x+9)+(y2-6y+9)+2(z2+2z+1)=0\(\Rightarrow\)(3x-3)2+(y-3)2+2(z+1)2=0\(\Rightarrow\hept{\begin{cases}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9x^2+y^2+2z^2-18x+4z-6z+20=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
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