\(2x^2-xy+9x-3y+4=0\)
\(\Rightarrow-y\left(x+3\right)+x\left(2x+9\right)=-4\)
\(\Rightarrow-y\left(x+3\right)=-4-x\left(2x+9\right)\)
\(\Rightarrow y=\dfrac{x\left(2x+9\right)+4}{x+3}=\dfrac{2x^2+9x+4}{x+3}\)
-Vì x,y nguyên nên:
\(\left(2x^2+9x+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(2x^2+6x+3x+9-5\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[2x\left(x+3\right)+3\left(x+3\right)-5\right]⋮\left(x+3\right)\)
\(\Rightarrow5⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;5;-1;-5\right\}\)
\(\Rightarrow x\in\left\{-2;2;-4;-8\right\}\)
*\(x=-2\Rightarrow y=\dfrac{2.\left(-2\right)^2+9.\left(-2\right)+4}{-2+3}=-6\)
\(x=2\Rightarrow y=\dfrac{2.2^2+9.2+4}{2+3}=6\)
\(x=-4\Rightarrow y=\dfrac{2.\left(-4\right)^2+9.\left(-4\right)+4}{-4+3}=0\)
\(x=-8\Rightarrow y=\dfrac{2.\left(-8\right)^2+9.\left(-8\right)+4}{-8+3}=-12\)
-Vậy các cặp số \(\left(x,y\right)\) là \(\left(-2,-6\right);\left(2,6\right);\left(-4,0\right);\left(-8;-12\right)\)
