$3x^2-2y^2-5xy +x-2y-7=0$
$\Leftrightarrow 3x^2 - 2y^2 - 6xy + xy + x - 2y - 7=0$
$\Leftrightarrow (3x^2 - 6xy) + (-2y^2 + xy) + ( x-2y)= 7$
$\Leftrightarrow 3x.(x-2y)+y. (x-2y)+(x-2y)= 7$
$\Leftrightarrow (x-2y)(3x+y+1)=7$
Xét các trường hợp :
$\text{TH1}:\Leftrightarrow \left[ \begin{array}{l}x-2y=1\\3x+y+1=7\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x-2y=1\\3x+y=6\end{array} \right.$
$\text{TH2}: \left[ \begin{array}{l}x-2y=7\\3x+y+1=1\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x-2y=7\\3x+y=0\end{array} \right.$
$\text{TH3}: \left[ \begin{array}{l}x-2y =-1\\3x+y+1=-7\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x-2y=-1\\3x+y=-8\end{array} \right.$
$\text{TH4}:\left[ \begin{array}{l}x-2y=-7\\3x+y+1=-1\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x-2y=-7\\3x+y=-2\end{array} \right.$
Từ đó suy ra các nghiệm $\in \mathbb{Z}$
Vậy $S=(1;- 3)$
