\(x+2xy-y=0\\ \Leftrightarrow x\left(2y+1\right)-y=0\\ \Leftrightarrow2x\left(2y+1\right)-2y-1+1=0\\ \Leftrightarrow2x\left(2y+1\right)-\left(2y+1\right)=-1\\ \Leftrightarrow\left(2x-1\right)\left(2y+1\right)=-1\)
Vì \(x,y\in N\Rightarrow\left\{{}\begin{matrix}2x-1\in Z,2y+1\in N\\2x-1,2y+1\inƯ\left(-1\right)\end{matrix}\right.\)
Ta có bảng:
2x-1 | -1 |
2y+1 | 1 |
x | 0 |
y | 0 |
Vậy \(\left(x,y\right)\in\left(0;0\right)\)