Giải:
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: \(2\left(n-1\right)⋮n-1\Rightarrow1⋮n-1\Rightarrow n-1\inƯ_{\left(1\right)}=\left\{\pm1\right\}.\)
Nếu:
\(\left[{}\begin{matrix}n-1=1\Rightarrow n=2\in Z.\\n-1=-1\Rightarrow n=0\in Z.\end{matrix}\right.\)
Vậy \(n\in\left\{0;2\right\}.\)