\(A=\dfrac{2n^2-4n+5}{n-2}=\dfrac{2n\left(n-2\right)+5}{n-2}=\dfrac{2n\left(n-2\right)}{n-2}+\dfrac{5}{n-2}=2+\dfrac{5}{n-2}\)
Ta có: \(2\) là số nguyên nên để \(A\) nguyên
Suy ra \(5⋮n-2\Rightarrow n-2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow n\in\left\{3;1;7\right\}\left(n\in N\text{*}\right)\)