Question

tìm các số nguyên dương x,y thỏa mãn:

2xy-5x+7y-4=0

Answer 1

\(2xy-5x+7y-4=0\)

\(\Leftrightarrow4xy-10x+14y-8=0\)

\(\Leftrightarrow2x\left(2y-5\right)+7\left(2y-5\right)+35-8=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2y-5\right)=-27\)

Mặt khác, do x, y nguyên dương nên \(\left\{{}\begin{matrix}2x+7>7\\2y-5>-5\end{matrix}\right.\)

Suy ra ta có 2 trường hợp:

• \(\left\{{}\begin{matrix}2x+7=27\\2y-5=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=2\end{matrix}\right.\)

• \(\left\{{}\begin{matrix}2x+7=9\\2y-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(10;2\right);\left(1;1\right)\)