\(\dfrac{x}{7}+\dfrac{1}{y}=\dfrac{-1}{14}\)
=>\(\dfrac{xy+7}{7y}=\dfrac{-1}{14}\)
=>\(\dfrac{2xy+14}{14y}=\dfrac{-y}{14y}\)
=>2xy+14=-y
=>2xy+y=-14
=>y(2x+1)=-14
mà 2x+1 lẻ(do x nguyên)
nên \(\left(2x+1\right)\cdot y=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)
=>\(\left(2x+1;y\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)