Đáp án:
\(n \in \left\{ {0;\,\,2} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \frac{{2n - 1}}{{n - 1}} = \frac{{2n - 2 + 1}}{{n - 1}} = 2 + \frac{1}{{n - 1}}\\
De\,\,\,P \in Z\,\,thi\,\,\,\frac{1}{{n - 1}} \in Z\\
\Rightarrow 1\,\,\, \vdots \,\,\left( {n - 1} \right) \Rightarrow n - 1 \in U\left( 1 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
n - 1 = 1\\
n - 1 = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
n = 2\,\,\,\left( {tm} \right)\\
n = 0\,\,\,\left( {tm} \right)
\end{array} \right.\\
Vay\,\,\,n \in \left\{ {0;\,\,2} \right\}.
\end{array}\)
