Đáp án:
d) \(\left[ \begin{array}{l}
n = - 7\\
n = 19\\
n = 5\\
n = 7
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{n + 4}}{{n + 1}} = \dfrac{{n + 1 + 3}}{{n + 1}} = 1 + \dfrac{3}{{n + 1}}\\
A \in Z \Leftrightarrow \dfrac{3}{{n + 1}} \in Z\\
\Leftrightarrow n + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
n + 1 = 3\\
n + 1 = - 3\\
n + 1 = 1\\
n + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 2\\
n = - 4\\
n = 0\\
n = - 2
\end{array} \right.\\
b)B = \dfrac{{n + 2}}{{n - 1}} = \dfrac{{n - 1 + 3}}{{n - 1}} = 1 + \dfrac{3}{{n - 1}}\\
B \in Z \Leftrightarrow \dfrac{3}{{n - 1}} \in Z\\
\Leftrightarrow n - 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
n - 1 = 3\\
n - 1 = - 3\\
n - 1 = 1\\
n - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 4\\
n = - 2\\
n = 2\\
n = 0
\end{array} \right.\\
c)C = \dfrac{{2n + 4}}{{n + 1}} = \dfrac{{2\left( {n + 1} \right) + 2}}{{n + 1}}\\
= 2 + \dfrac{2}{{n + 1}}\\
C \in Z\\
\to \dfrac{2}{{n + 1}} \in Z\\
\to n + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
n + 1 = 2\\
n + 1 = - 2\\
n + 1 = 1\\
n + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 1\\
n = - 3\\
n = 0\\
n = - 2
\end{array} \right.\\
d)D = \dfrac{{2n + 1}}{{6 - n}} = \dfrac{{ - 2\left( {6 - n} \right) + 13}}{{6 - n}}\\
= - 2 + \dfrac{{13}}{{6 - n}}\\
D \in Z \Leftrightarrow \dfrac{{13}}{{6 - n}} \in Z\\
\to 6 - n \in U\left( {13} \right)\\
\to \left[ \begin{array}{l}
6 - n = 13\\
6 - n = - 13\\
6 - n = 1\\
6 - n = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = - 7\\
n = 19\\
n = 5\\
n = 7
\end{array} \right.
\end{array}\)
