\(xy+3x-5y-15+7=0\Leftrightarrow x\left(y+3\right)-5\left(y+3\right)=-7\)
\(\Leftrightarrow\left(x-5\right)\left(y+3\right)=-7\)
Ta có các trường hợp sau:
TH1: \(\left\{{}\begin{matrix}x-5=-1\\y+3=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x-5=1\\y+3=-7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=-10\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x-5=7\\y+3=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=12\\y=-4\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x-5=-7\\y+3=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-2\end{matrix}\right.\)
Vậy pt có 4 cặp nghiệm nguyên:
\(\left(x;y\right)=\left(4;4\right);\left(6;-10\right);\left(12;-4\right);\left(-2;-2\right)\)
