\(P=\dfrac{2n-1}{n-1}\) nguyên \(\Leftrightarrow\) \(2n-1⋮n-1\) \(\Rightarrow2n-1⋮2n-2\)
ta có : \(2n-1=2n-2+1\)
\(\Rightarrow\) \(1⋮n-1\) vậy \(n-1\) là ước của \(1\) là \(\pm1\)
\(n-1=1\Leftrightarrow n=2\)
\(n-1=-1\Leftrightarrow n=0\)
vậy \(n=2;n=0\) nguyên thì \(\dfrac{2n-1}{n-1}\) nguyên
Để \(p=\dfrac{2n-1}{n-1}\in Z\) thì \(2n-1⋮n-1\)
\(\Rightarrow2n-2+1⋮n-1\)
\(\Rightarrow1⋮n-1\left(Do2n-2⋮n-1\right)\)
\(\Rightarrow n-1\inƯ\left(1\right)\)
\(\Rightarrow n-1\in\left\{\pm1\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\\n-1=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=2\\n=0\end{matrix}\right.\)
Vậy \(n\in\left\{2;0\right\}\)
