a) (x-3).(2y+1)=7 có \(7=1.7=-1.-7\) nên ta có các trường hợp sau:
∙ Nếu \(\left\{\begin{matrix}x-3=1\\3y+1=7\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=4\\y=3\end{matrix}\right.\)
∙ Nếu \(\left\{\begin{matrix}x-3=7\\2y+1=1\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=10\\y=0\end{matrix}\right.\)
∙ Nếu \(\left\{\begin{matrix}x-3=-1\\2y+1=-7\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=2\\y=-4\end{matrix}\right.\)
∙ Nếu \(\left\{\begin{matrix}x-3=-7\\2y+1=-1\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-4\\y=-1\end{matrix}\right.\)
Vậy: \(\left(x;y\right)=\left(4;3\right),\left(10;0\right),\left(2;-4\right),\left(-4;-1\right)\)
b) (2x+1).(3y-2)=-55
\(\Rightarrow2x+1=-\frac{55}{3y-2}\left(1\right)\)
Để \(x\in Z\) thì \(3y-2\inƯ_{\left(55\right)}=-55;-11;-5;-1;1;5;11;55\)
* \(3y-2=55\Rightarrow3y=57\Rightarrow y=19\) thay vào \(\left(1\right)\Rightarrow x=0\)
* \(3y-2=11\Rightarrow3y=13\Rightarrow y=\frac{13}{3}\) ( loại )
* \(3y-2=5\Rightarrow3y=7\Rightarrow y=\frac{7}{3}\) ( loại )
* \(3y-2=1\Rightarrow3y=3\Rightarrow y=1\) thay vào \(\left(1\right)\Rightarrow x=-28\)
* \(3y-2=-1\Rightarrow3y=1\Rightarrow y=\frac{1}{3}\) ( loại )
* \(3y-2=-5\Rightarrow3y=-3\Rightarrow y=-1\) thay vào \(\left(1\right)\Rightarrow x=-6\) * \(3y-2=-11\Rightarrow3y=-9\Rightarrow y=-3\) thay vào \(\left(1\right)\Rightarrow x=-28\) * \(3y-2=-55\Rightarrow3y=-53\Rightarrow y=-\frac{53}{3}\) ( loại ) Vậy: Ta có 4 cặp số \(\left(x;y\right)\) thỏa mãn là: \(\left(x;y\right)=\left(0;19\right),\left(-28;1\right),\left(-6;-1\right),\left(-28;-3\right)\)
