Với \(n\ne3.\)
\(A=\dfrac{2n-1}{n-3}=\dfrac{2n-6+5}{n-3}=2+\dfrac{5}{n-3}\)
\(A\) nguyên \(\Rightarrow\dfrac{5}{n-3}\) nguyên.
\(\Rightarrow5⋮n-3.\)
\(n\in Z\Rightarrow n-3\in Z\)
\(\Rightarrow n-3\inƯ\left(5\right)\)
\(\Rightarrow n-3\in\left\{\pm1;\pm5\right\}\)
\(n-3\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
\(n\) | \(4\) | \(2\) | \(8\) | \(-2\) |
Vậy \(n\in\left\{4;2;8;-2\right\}.\)