Ta có: (2x+1)(y-3)=2
nên 2x+1;y-3 là các ước của 2
Trường hợp 1:
\(\left\{{}\begin{matrix}2x+1=1\\y-3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=5\end{matrix}\right.\)
Trường hợp 2:
\(\left\{{}\begin{matrix}2x+1=-1\\y-3=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-2\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Trường hợp 3:
\(\left\{{}\begin{matrix}2x+1=2\\y-3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=4\end{matrix}\right.\)
Trường hợp 4:
\(\left\{{}\begin{matrix}2x+1=-2\\y-3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-3\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=2\end{matrix}\right.\)
Vậy: \(\left(x,y\right)=\left\{\left(0;5\right);\left(-1;1\right);\left(\dfrac{1}{2};4\right);\left(\dfrac{-3}{2};2\right)\right\}\)
(2x+1)(y-3)=2
=> \(2x+1vày-3\inƯ\left(2\right)\)
Ta có: \(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)
Vậy....
