\(x+2y=3xy+3\)
\(\Leftrightarrow3xy-x-2y+3=0\)
\(\Leftrightarrow9xy-3x-6y+9=0\)
\(\Leftrightarrow3x\left(3y-1\right)-2\left(3y-1\right)+7=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3y-1\right)=-7\)
3x-2 | -7 | -1 | 1 | 7 |
3y-1 | 1 | 7 | -7 | -1 |
x | -5/3(ktm) | 1/3(ktm) | 1 | 3 |
y | 2/3(ktm) | 8/3(ktm) | -2 | 0 |
Vậy \(\left(x;y\right)=\left(1;-2\right);\left(3;0\right)\)