\(\left(2x+1\right)\left(y-3\right)=10\)
\(\Leftrightarrow2x+1;y-3\inƯ\left(10\right)\)
Mà \(x,y\in N\Leftrightarrow2x+1;y-3\in N\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+1=1\\y-3=10\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=2\\y-3=5\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=10\\y-3=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=5\\y-3=2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=12\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=8\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{9}{2}\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\end{matrix}\right.\)
Vì \(\left(2x+1\right)\left(y-3\right)=10\)
\(\Rightarrow\left\{{}\begin{matrix}2x+1\\y-3\end{matrix}\right.\inƯ\left(10\right)\)
Ta có: \(Ư\left(10\right)=\left\{1;2;5;10\right\}\)
Nếu:
+)\(\left\{{}\begin{matrix}2x+1=1\Leftrightarrow x=0\\y-3=10\Leftrightarrow y=13\end{matrix}\right.\)
+)\(\left\{{}\begin{matrix}2x+1=2\Leftrightarrow x=0,5\\y-3=5\Leftrightarrow y=8\end{matrix}\right.\)(loại, vì \(x\notin N\))
+)\(\left\{{}\begin{matrix}2x+1=5\Leftrightarrow x=2\\y-3=2\Leftrightarrow y=5\end{matrix}\right.\)
+)\(\left\{{}\begin{matrix}2x+1=10\Leftrightarrow x=4,5\\y-3=1\Leftrightarrow y=4\end{matrix}\right.\)(loại, vì \(x\notin N\))
Vậy \(\left(x;y\right)\in\left\{\left(0;13\right),\left(2;5\right)\right\}\)
Có 2 cặp (x;y)
