\(P=\dfrac{2n-1}{n-1}=\dfrac{2n-2+1}{n-1}=2+\dfrac{1}{n-1}\)
Do P nguyên, 2 nguyên \(\Rightarrow\dfrac{1}{n-1}\) nguyên
\(\Rightarrow1⋮\left(n-1\right)\Rightarrow n-1=Ư\left(1\right)=\left\{-1;1\right\}\)
\(n-1=-1\Rightarrow n=0\)
\(n-1=1\Rightarrow n=2\)
Vậy \(n=\left\{0;2\right\}\)