Đáp án:
\(\left[ \begin{array}{l}
n = 46\\
n = 5\\
n = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A \in Z\\
\to n + 3 \vdots 2n - 2\\
\to 2n + 6 \vdots 2n - 2\\
\to 2n - 2 + 8 \vdots 2n - 2\\
\to 8 \vdots 2n - 2\\
\to 4 \vdots n - 1\\
\to 2n - 2 \in U\left( 8 \right)\\
\to \left[ \begin{array}{l}
2n - 2 = 8\\
2n - 2 = - 8\\
2n - 2 = 4\\
2n - 2 = - 4\\
2n - 2 = 2\\
2n - 2 = - 2\\
2n - 2 = 1\\
2n - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 5\\
n = - 3\\
n = 3\\
n = - 1\\
n = 2\\
n = 0\\
n = \frac{3}{2}\left( l \right)\\
n = \frac{1}{2}\left( l \right)
\end{array} \right.\\
b.12 \vdots 3n - 1\\
\to 3n - 1 \in U\left( {12} \right)\\
\to \left[ \begin{array}{l}
3n - 1 = 12\\
3n - 1 = - 12\\
3n - 1 = 6\\
3n - 1 = - 6\\
3n - 1 = 4\\
3n - 1 = - 4\\
3n - 1 = 3\\
3n - 1 = - 3\\
3n - 1 = 2\\
3n - 1 = - 2\\
3n - 1 = 1\\
3n - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = \frac{{13}}{3}\left( l \right)\\
n = - \frac{{11}}{3}\left( l \right)\\
n = \frac{7}{3}\left( l \right)\\
n = - \frac{5}{3}\left( l \right)\\
n = \frac{5}{3}\left( l \right)\\
n = - 1\left( {TM} \right)\\
n = \frac{4}{3}\left( l \right)\\
n = - \frac{2}{3}\left( l \right)\\
n = 1\left( {TM} \right)\\
n = - \frac{1}{3}\left( l \right)\\
n = \frac{2}{3}\left( l \right)\\
n = 0\left( {TM} \right)
\end{array} \right.\\
c.2n + 3 \vdots 7\\
\to \left( {2n + 3 - 7} \right) \vdots 7\\
\to 2n - 4 \vdots 7\\
\to 2\left( {n - 2} \right) \vdots 7\\
Mà:\left( {2;7} \right) = 1\\
\to \left( {n - 2} \right) \vdots 7\\
\to n - 2 = 7k\left( {k \in Z} \right)\\
\to n = 2 + 7k
\end{array}\)
Vậy với \(n = 2 + 7k\) thì \(\frac{{2n + 3}}{7}\) nguyên
\(\begin{array}{l}
d.D = \frac{{8n + 193}}{{4n + 3}} = \frac{{2\left( {4n + 3} \right) + 187}}{{4n + 3}} = 2 + \frac{{187}}{{4n + 3}}\\
D \in Z\\
\Leftrightarrow \frac{{187}}{{4n + 3}} \in Z\\
\to 4n + 3 \in U\left( {187} \right)\\
\to \left[ \begin{array}{l}
4n + 3 = 187\\
4n + 3 = - 187\\
4n + 3 = 17\\
4n + 3 = - 17\\
4n + 3 = 11\\
4n + 3 = - 11\\
4n + 3 = 1\\
4n + 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 46\\
n = - \frac{{95}}{2}\left( l \right)\\
n = 5\\
n = - \frac{{14}}{4}\left( l \right)\\
n = - \frac{1}{2}\left( l \right)\\
n = - 1
\end{array} \right.
\end{array}\)
