Ta có:\(3x^2-18y^2+2z^2+3y^2z^2-18x=27\)
\(\Leftrightarrow3x^2-18y^2+2z^2+3y^2z^2-18x-27=0\)
\(\Leftrightarrow3\left(x^2-6x+9\right)-18y^2+2z^2+3y^2z^2-54=0\)
\(\Leftrightarrow3\left(x-3\right)^2-18y^2+2z^2+3y^2z^2=54\)
Để pt có nghiệm nguyên thì:\(z^2⋮3\) \(\Rightarrow z⋮3\)\(\Rightarrow z^2⋮9\)\(\Rightarrow z^2\ge9\)
\(\Leftrightarrow3\left(x-3\right)^2+3y^2\left(z^2-6\right)+2z^2=54\)
\(\Rightarrow54=3\left(x-3\right)^2+3y^2\left(z^2-6\right)+2z^2\ge3\left(x-3\right)^2\le12\)
\(\Rightarrow y^2\le4\Rightarrow\hept{\begin{cases}y^2=1\\y^2=4\end{cases}}\)
Với \(y^2=1\Rightarrow y=1\)pt có dạng :
\(3\left(x-3\right)^2+5z^2=72\)
\(\Leftrightarrow5z^2\le72\)
\(\Leftrightarrow z^2=9\Leftrightarrow z=3\)
\(\Rightarrow x=6\)
Với \(y^2=4\Rightarrow y=2\)pt có dạng:
\(3\left(x-3\right)^2+14z^2=126\)
\(\Leftrightarrow14z^2\le126\)
\(\Leftrightarrow z^2\le9\Rightarrow z=3\)
\(\Rightarrow x=3\)
Vậy ......