Ta có 2x+3y+xy=11
\(\Rightarrow\)x(y+2)+3y=11
\(\Rightarrow\)x(y+2)+3y+6=11+6
\(\Rightarrow\)x(y+2)+3(y+2)=17
\(\Rightarrow\)(x+3)(y+2)=17
\(\Rightarrow\)x+3,y+2\(\in\)Ư(17)
do x,y\(\in\)Z \(\Rightarrow\)x,y\(\in\left\{\left(-20,-3\right);\left(-4,-19\right);\left(-2,15\right)\left(14,-1\right)\right\}\)