3xy-5x-6y+7=0
=>3xy-6y-5x+10-3=0
=>3y(x-2)-5(x-2)=3
=>(x-2)(3y-5)=3
=>\(\left(x-2\right)\left(3y-5\right)=1\cdot3=3\cdot1=\left(-1\right)\cdot\left(-3\right)=\left(-3\right)\cdot\left(-1\right)\)
=>\(\left(x-2;3y-5\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(3;\dfrac{8}{3}\right);\left(5;2\right);\left(1;\dfrac{2}{3}\right);\left(-1;\dfrac{4}{3}\right)\right\}\)
mà x,y nguyên
nên x=5;y=2
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