a: \(2x^2-5x+2xy-5y-3=7\)
=>\(x\left(2x-5\right)+y\cdot\left(2x-5\right)=10\)
=>(2x-5)(x+y)=10
mà 2x-5 lẻ
nên \(\left(2x-5\right)\left(x+y\right)=1\cdot10=\left(-1\right)\cdot\left(-10\right)=5\cdot2=\left(-5\right)\cdot\left(-2\right)\)
=>\(\left(2x-5;x+y\right)\in\left\{\left(1;10\right);\left(-1;-10\right);\left(5;2\right);\left(-5;-2\right)\right\}\)
=>\(\left(x;x+y\right)\in\left\{\left(3;10\right);\left(2;-10\right);\left(5;2\right);\left(0;-2\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(3;7\right);\left(2;-12\right);\left(5;-3\right);\left(0;-2\right)\right\}\)
b: \(2xy-3y^2+4x-6y-2=5\)
=>\(y\left(2x-3y\right)+2\left(2x-3y\right)=7\)
=>(2x-3y)(y+2)=7
=>\(\left(2x-3y;y+2\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
=>\(\left(2x-3y;y\right)\in\left\{\left(1;5\right);\left(7;-1\right);\left(-1;-9\right);\left(-7;-3\right)\right\}\)
=>\(\left(2x;y\right)\in\left\{\left(16;5\right);\left(4;-1\right);\left(-28;-9\right);\left(-16;-3\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(8;5\right);\left(2;-1\right);\left(-14;-9\right);\left(-8;-3\right)\right\}\)
