Ta có : 9x2 + y2 + 2z2 - 18x + 4z - 6y + 20 = 0
<=> 9x2 - 18x + 9 + y2 - 6y + 9 + 2z2 + 4z + 2 = 0
<=> 9(x2 - 2x + 1) + (y2 - 6y + 9) + 2(z2 + 2z + 1) = 0
<=> 9(x - 1)2 + (y - 3)2 + 2(z + 1)2 = 0 (*)
Vì \(9\left(x-1\right)^2\ge0\forall x\in R\)
\(\left(y-3\right)^2\ge0\forall y\in R\)
\(2\left(z+1\right)^2\ge0\forall z\in R\)
Nên : pt (*) <=> \(\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
Vậy pt có nhiệm (x;y;z) = (1;3;-1)
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-3)^2+2(z^2+2z+1)=0 <=>9(x-1)^2+(y-3)^2+2(z+1)^2=0 <=>x=1;y=3;z=-1
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)(1)
\(\text{Do: }\left(x-1\right)^2\ge0\)
\(\left(y-3\right)^2\ge0\)
\(\left(z+1\right)^2\ge0\)
\(\text{Nen: }\left(1\right)\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
\(\Rightarrow\left(x;y;z\right)=\left(\pm1;3\right)\)
9x2+y2+2z2-18x+4z-6y+20=0
=> (3x)2 - 2.3x.3 + 32 + y2 - 2.y.3 + 32 + 2z2 - 4z + 2 = 0
=> (3x - 3)2 + (y - 3)2 + 2z(z - 2) + 2 = 0
=> (3x - 3)2 + (y - 3)2 + 2z(z - 2) = -2
Mà \(\hept{\begin{cases}\left(3x-3\right)^2\ge0\\\left(y-3\right)^2\ge0\end{cases}}\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2\ge0\)
\(\Rightarrow2z\left(z-2\right)=-2\Rightarrow z=1\)
=> x = 1, y = 3, z = 1
