Đáp án:
\(\left[ \begin{array}{l}
n = 1\\
n = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \dfrac{1}{2}\\
A = \dfrac{{2n + 15}}{{2n - 1}} = \dfrac{{2n - 1 + 16}}{{2n - 1}}\\
= 1 + \dfrac{{16}}{{2n - 1}}\\
Để:A \in Z\\
\to \dfrac{{16}}{{2n - 1}} \in Z\\
\to 2n - 1 \in U\left( {16} \right)\\
\to \left[ \begin{array}{l}
2n - 1 = 16\\
2n - 1 = - 16\\
2n - 1 = 8\\
2n - 1 = - 8\\
2n - 1 = 4\\
2n - 1 = - 4\\
2n - 1 = 2\\
2n - 1 = - 2\\
2n - 1 = 1\\
2n - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = \dfrac{{17}}{2}\left( l \right)\\
n = - \dfrac{{15}}{2}\left( l \right)\\
n = \dfrac{9}{2}\left( l \right)\\
n = - \dfrac{7}{2}\left( l \right)\\
n = \dfrac{5}{2}\left( l \right)\\
n = - \dfrac{3}{2}\left( l \right)\\
n = \dfrac{3}{2}\left( l \right)\\
n = - \dfrac{1}{2}\\
n = 1\left( {TM} \right)\\
n = 0\left( {TM} \right)
\end{array} \right.\\
KL:\left[ \begin{array}{l}
n = 1\\
n = 0
\end{array} \right.
\end{array}\)
