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3xy+2x-5y=6

=>\(x\left(3y+2\right)-5y-\dfrac{10}{3}=6-\dfrac{10}{3}=\dfrac{8}{3}\)

=>\(3x\left(y+\dfrac{2}{3}\right)-5\left(y+\dfrac{2}{3}\right)=\dfrac{8}{3}\)

=>\(\left(y+\dfrac{2}{3}\right)\left(3x-5\right)=\dfrac{8}{3}\)

=>\(\left(3x-5\right)\left(3y+2\right)=8\)

=>\(\left(3x-5\right)\left(3y+2\right)=1\cdot8=8\cdot1=\left(-1\right)\cdot\left(-8\right)=\left(-8\right)\cdot\left(-1\right)=2\cdot4=4\cdot2=\left(-2\right)\cdot\left(-4\right)=\left(-4\right)\cdot\left(-2\right)\)

=>\(\left(3x-5;3y+2\right)\in\left\{\left(1;8\right);\left(8;1\right);\left(-1;-8\right);\left(-8;-1\right);\left(2;4\right);\left(4;2\right);\left(-2;-4\right);\left(-4;-2\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(2;2\right);\left(\dfrac{13}{3};-\dfrac{1}{3}\right);\left(\dfrac{4}{3};-\dfrac{10}{3}\right);\left(-1;-1\right);\left(\dfrac{7}{3};\dfrac{2}{3}\right);\left(3;0\right);\left(1;-2\right);\left(\dfrac{1}{3};-\dfrac{4}{3}\right)\right\}\)

mà (x,y) nguyên

nên \(\left(x;y\right)\in\left\{\left(2;2\right);\left(-1;-1\right);\left(3;0\right);\left(1;-2\right)\right\}\)