BÀi 1: 2xy-x-y=1
=>x(2y-1)-y=1
=>\(2x\left(y-\frac12\right)-y+\frac12=\frac32\)
=>\(\left(2x-1\right)\left(y-\frac12\right)=\frac32\)
=>(2x-1)(2y-1)=3
=>(2x-1;2y-1)∈{(1;3);(3;1);(-1;-3);(-3;-1)}
=>(2x;2y)∈{(2;4);(4;2);(0;-2);(-2;0)}
=>(x;y)∈{(1;2);(2;1);(0;-1);(-1;0)}


