a: 6xy+3y+4x-8=0
=>3y(2x+1)+4x+2-10=0
=>(2x+1)(3y+2)=10
mà 2x+1 lẻ và 3y+2 chia 3 dư 2(Do x,y là các số nguyên)
nên \(\left(2x+1\right)\left(3y+2\right)=1\cdot10=5\cdot2=\left(-1\right)\cdot\left(-10\right)=\left(-5\right)\cdot\left(-2\right)\)
=>\(\left(2x+1;3y+2\right)\in\left\{\left(1;10\right);\left(5;2\right);\left(-1;-10\right);\left(-5;-2\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;\dfrac{8}{3}\right);\left(2;0\right);\left(-1;-4\right);\left(-3;-\dfrac{4}{3}\right)\right\}\)
mà x,y nguyên
nên \(\left(x;y\right)\in\left\{\left(2;0\right);\left(-1;-4\right)\right\}\)
b: \(\dfrac{x}{3}+\dfrac{-1}{-4}=\dfrac{-1}{y-8}\)
=>\(\dfrac{x}{3}+\dfrac{1}{4}=-\dfrac{1}{y-8}\)
=>\(\dfrac{4x+3}{12}=\dfrac{-1}{y-8}\)
=>\(\left(4x+3\right)\left(y-8\right)=-12\)
mà 4x+3 chia 4 dư 3
nên \(\left(4x+3\right)\left(y-8\right)=\left(-1\right)\cdot12=3\cdot\left(-4\right)\)
=>\(\left(x;y\right)\in\left\{\left(-1;20\right);\left(0;4\right)\right\}\)