\(\Leftrightarrow x^2+y^2+2z^2-xy-2yz+2xz+3x-3y+8< 0\)
\(\Leftrightarrow\left(x^2+\dfrac{1}{4}y^2+z^2+\dfrac{9}{4}-xy-yz+2xz+3x-\dfrac{3}{2}y+3z\right)+\left(\dfrac{1}{4}y^2+z^2+\dfrac{9}{4}-yz+\dfrac{3}{2}y-3z\right)+\dfrac{1}{2}\left(y^2-6y+9\right)-1< 0\)
\(\Leftrightarrow\left(x-\dfrac{y}{2}+z+\dfrac{3}{2}\right)^2+\left(\dfrac{y}{2}-z+\dfrac{3}{2}\right)^2+\dfrac{1}{2}\left(y-3\right)^2< 1\) (1)
\(\Rightarrow\dfrac{1}{2}\left(y-3\right)^2< 1\)
\(\Rightarrow\left(y-3\right)^2< 2\Rightarrow\left[{}\begin{matrix}\left(y-3\right)^2=0\\\left(y-3\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=3\\y=4\\y=2\end{matrix}\right.\)
- Với \(y=2\) thế vào (1):
\(\left(x+z+\dfrac{1}{2}\right)^2+\left(z-\dfrac{5}{2}\right)^2+\dfrac{1}{2}< 1\)
\(\Leftrightarrow\left(x+z+\dfrac{1}{2}\right)^2+\left(z-\dfrac{5}{2}\right)^2< \dfrac{1}{2}\)
\(\Rightarrow\left(z-\dfrac{5}{2}\right)^2< \dfrac{1}{2}\Rightarrow\left(z-\dfrac{5}{2}\right)^2=0\)
\(\Rightarrow z=\dfrac{5}{2}\notin Z\) (loại)
- Với \(y=3\) thế vào (1):
\(\left(x+z\right)^2+\left(z-3\right)^2< 1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\z=3\end{matrix}\right.\)
- Với \(y=4\) thế vào (1)
\(\left(x+z-\dfrac{1}{2}\right)^2+\left(z-\dfrac{7}{2}\right)^2< \dfrac{1}{2}\)
\(\Rightarrow\left(z-\dfrac{7}{2}\right)^2< \dfrac{1}{2}\Rightarrow\left(z-\dfrac{7}{2}\right)^2=0\)
\(\Rightarrow z=\dfrac{7}{2}\notin Z\) (loại)
Vậy \(\left(x;y;z\right)=\left(-3;3;3\right)\)
