\(\Leftrightarrow x\left(y+1\right)+y+1=31\\ \Leftrightarrow\left(x+1\right)\left(y+1\right)=31=31\cdot1\)
\(TH_1:\left\{{}\begin{matrix}x+1=31\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=0\end{matrix}\right.\rightarrow\left(30;0\right)\\ TH_2:\left\{{}\begin{matrix}x+1=1\\y+1=31\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=30\end{matrix}\right.\rightarrow\left(0;30\right)\)
Vậy \(\left(x;y\right)\in\left\{\left(30;0\right);\left(0;30\right)\right\}\)
\(\Leftrightarrow xy+x+y+1=31\Leftrightarrow\left(x+1\right)\left(y+1\right)=31\)
Do x, y nguyên dương nên xảy ra các TH sau:
TH1: \(\left\{{}\begin{matrix}x+1=1\\y+1=31\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=30\end{matrix}\right.\)(LOẠI)
TH2:\(\left\{{}\begin{matrix}x+1=31\\y+1=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=0\end{matrix}\right.\)(LOẠI)
Vậy không có x, y nguyên dương thoả mãn đề bài.
