\(A=\dfrac{x^2+x-6+7}{x-2}=\dfrac{\left(x-2\right)\left(x+3\right)+7}{x-2}=x+3+\dfrac{7}{x-2}\)
Do \(x+3\) nguyên khi x nguyên nên để A nguyên thì \(\dfrac{7}{x-2}\) nguyên
\(\Rightarrow x-2\inƯ\left(7\right)\)
\(\Rightarrow x-2\in\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x\in\left\{-5;1;3;9\right\}\)