Đáp án:
c. \(\left[ \begin{array}{l}
n = 9\\
n = 3\\
n = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.6 \vdots n - 1\\
\Leftrightarrow n - 1 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
n - 1 = 6\\
n - 1 = 3\\
n - 1 = 2\\
n - 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
n = 7\\
n = 4\\
n = 3\\
n = 2
\end{array} \right.\\
b.15 \vdots 2n - 1\\
\Leftrightarrow 2n - 1 \in U\left( {15} \right)\\
\to \left[ \begin{array}{l}
2n - 1 = 15\\
2n - 1 = 5\\
2n - 1 = 3\\
2n - 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
2n = 16\\
2n = 6\\
2n = 4\\
2n = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
n = 8\\
n = 3\\
n = 2\\
n = 1
\end{array} \right.\\
c.n + 9 \vdots n\\
\Leftrightarrow 9 \vdots n\\
\Leftrightarrow n \in U\left( 9 \right)\\
\to \left[ \begin{array}{l}
n = 9\\
n = 3\\
n = 1
\end{array} \right.\\
d.5n - 6 \vdots n\\
\Leftrightarrow 6 \vdots n\\
\to n \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
n = 6\\
n = 3\\
n = 2
\end{array} \right.\left( {Do:n > 1} \right)
\end{array}\)
