nếu A nguyên thì .-. .........
ta có :
\(A=\dfrac{2n-1}{n+2}=\dfrac{2n+4-5}{n+2}=\dfrac{2\left(n+2\right)-5}{n+2}=\dfrac{2\left(n+2\right)}{n+2}-\dfrac{5}{n+2}=2-\dfrac{5}{n+2}\\ \text{đ}\text{ể }Anguy\text{ê}n.th\text{ì }2-\dfrac{5}{n+2}nguy\text{ê}n\\ \Rightarrow\dfrac{5}{n-2}nguy\text{ê}n\\ \Rightarrow5⋮n-2\\ \Rightarrow n-2\in\text{Ư}\left(5\right)=\left\{-1;1;5;-5\right\}\)
ta có bảng :
| n-2 | 1 | -1 | 5 | -5 |
| n | 3 | 1 | 7 | -3 |
vậy ...
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