Đáp án:
d) \(\left[ \begin{array}{l}
n = 2\\
n = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)3n + 7 \vdots n\\
\Leftrightarrow 7 \vdots n\\
\Leftrightarrow n \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
n = 7\\
n = 1
\end{array} \right.\\
b)27 - 5n \vdots n + 1\\
\to - 5\left( {n + 1} \right) + 32 \vdots n + 1\\
\to 32 \vdots n + 1\\
\to \left[ \begin{array}{l}
n + 1 = 32\\
n + 1 = 16\\
n + 1 = 8\\
n + 1 = 4\\
n + 1 = 2\\
n + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 31\\
n = 15\\
n = 7\\
n = 3\\
n = 1\\
n = 0
\end{array} \right.\\
c)3n + 1 \vdots 11 - 2n\\
\to - \left( { - 3n - 1} \right) \vdots 11 - 2n\\
\to - \left( { - 6n - 2} \right) \vdots 11 - 2n\\
\to - 3\left( { - 2n + 11} \right) + 35 \vdots 11 - 2n\\
\to 35 \vdots 11 - 2n\\
\to 11 - 2n \in U\left( {35} \right)\\
\to \left[ \begin{array}{l}
11 - 2n = 35\\
11 - 2n = 7\\
11 - 2n = 5\\
11 - 2n = 1
\end{array} \right. \to \left[ \begin{array}{l}
n = - 12\\
n = 2\\
n = 3\\
n = 5
\end{array} \right.\\
d)3n + 4 \vdots 2n + 1\\
\to 6n + 8 \vdots 2n + 1\\
\to 3\left( {2n + 1} \right) + 5 \vdots 2n + 1\\
\to 5 \vdots 2n + 1\\
\to 2n + 1 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
2n + 1 = 5\\
2n + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
n = 2\\
n = 0
\end{array} \right.
\end{array}\)
