\(x^2+y^2+z^2=4x-2y+6z-14\\ \Leftrightarrow x^2+y^2+x^2-4x+2y-6z+4+1+9=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+\left(z^2-6z+9\right)=0\\ \Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)
Ta có :
\(\left(x-2\right)^2\ge0\forall x\\ \left(y+1\right)^2\ge0\forall y\\ \left(z-3\right)^2\ge0\forall z\)
\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)
Dấu " = " xảy ra khi :
\(\left[{}\begin{matrix}x-2=0\\y+1=0\\z-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\y=-1\\z=3\end{matrix}\right.\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y-1\right)^2+\left(z-3\right)^2=0\)
\(\Leftrightarrow........\)
