Ta có: \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{y+5}\)
=>\(\dfrac{2x-3}{6}=\dfrac{1}{y+5}\)
=>\(\left(2x-3\right)\left(y+5\right)=6\)
mà 2x-3 lẻ
nên \(\left(2x-3;y+5\right)\in\left\{\left(1;6\right);\left(-1;-6\right);\left(3;2\right);\left(-3;-2\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(2;1\right);\left(1;-11\right);\left(3;-3\right);\left(0;-7\right)\right\}\)