\(A=\frac{2n-1}{n+3}=\frac{2n+6-7}{n+3}=\frac{2\left(n+3\right)-7}{n+3}=\frac{2\left(n+3\right)}{n+3}-\frac{7}{n+3}=2+\frac{7}{n+3}\)
A nguyên <=>\(2+\frac{7}{n+3}\) nguyên
<=>7 chia hết cho n+3
<=>\(n+3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
<=>\(n\in\left\{-10;-4;-2;4\right\}\)
Vậy A nguyên khi \(n\in\left\{-10;-4;-2;4\right\}\)