Ta có:

\(x^2y+5y-x+2=0\)

\(\Rightarrow y\left(x^2+5\right)=x-2\)

\(\Rightarrow y=\dfrac{x-2}{x^2+5}\left(1\right)\)

Để y nhận giá trị nguyên (y∈Z) thì (x - 2) ⋮ (x\(^2\) + 5)

\(\Rightarrow\left(x-2\right)\vdots\left(x^2+5\right)\)

\(\Rightarrow\left(x-2\right)\left(x+2\right)\vdots\left(x^2+5\right)\)

\(\Rightarrow\left(x^2-4\right)\vdots\left(x^2+5\right)\)

\(\Rightarrow\left(x^2+5-9\right)\vdots\left(x^2+5\right)\)

Mà \(\left(x^2+5-9\right)\vdots x^2+5\)

\(\Rightarrow9\vdots\left(x^2+5\right)\)

Hay \(\left(x^2+5\right)\inƯ\left(9\right)\)

Mà \(x^2+5\ge5\forall x\in\R\)

\(\Rightarrow x^2+5=9\)

\(\Rightarrow x^2=4\)

\(\Rightarrow\left[\begin{array}{l}x=2\\ x=-2\end{array}\right.\)

Với \(x=2\Rightarrow y=\dfrac{2-2}{2^2+5}=\dfrac09=0\left(t/m\right)\)

Với \(x=-2\Rightarrow y=\dfrac{-2-2}{\left(-2\right)^2+5}=\dfrac{-4}{9}\left(loại\right)\)

Vậy \(\left(x;y\right)=\left(2;0\right)\)

Chúc mừng bạn nhé!