\(x^2+2y^2+3xy-x-y+3=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+2y\right)-\left(x+y\right)=-3\)
\(\Leftrightarrow\left(x+y\right)\left(x+2y-1\right)=-3\)
Ta có bảng:
x+y | -3 | -1 | 1 | 3 |
x+2y-1 | 1 | 3 | -3 | -1 |
x | -4 | -6 | 4 | 6 |
y | 3 | 5 | -3 | -3 |
Vậy \(\left(x;y\right)=\left(-4;3\right);\left(-6;5\right);\left(4;-3\right);\left(6;-3\right)\)