\(xy^2\) - \(2xy\) + \(x\) + \(y^2\) = 6
\(x\)( \(y^2\) - \(2y\) + 1 ) + \(y^2\) - 1 = 5
\(x\) ( \(y-1\) ) 2 + ( \(y-1\))(\(y+1\)) = 5
(\(y-1\))( \(xy-x\) + y + 1) = 5
Ư(5) ={ -5; -1; 1; 5)
ta có bảng :
y- 1 | -5 | -1 | 1 | 5 |
y | -4 | 0 | 2 | 6 |
xy-x+y+1 | -1 | -5 | 5 | 1 |
x | -2/5 | 6 | 2 | -6/5 |
Vì x, y \(\in\) Z nên (x, y ) = ( 0; 6); ( 2; 2)