Ta có: \(P=\dfrac{2n-1}{n-1}=\dfrac{2n-2+1}{n-1}=\dfrac{2\left(n-1\right)+1}{n-1}=\dfrac{2\left(n-1\right)}{n-1}+\dfrac{1}{n-1}\)
Để P là số nguyên thì: \(2\left(n-1\right)⋮n-1\)
\(\Rightarrow1⋮n-1\)
\(\Rightarrow n-1\inƯ\left(1\right)=\left\{-1;1\right\}\)
+ Nếu: \(n-1=1\Rightarrow n=2\in Z\)
+ Nếu: \(n-1=-1\Rightarrow n=0\in Z\)
Vậy \(n\in\left\{2;0\right\}\)