Đáp án:
d) \(\left[ \begin{array}{l}
n = 2\\
n = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{8n + 11}}{{2n + 1}} = \dfrac{{4\left( {2n + 1} \right) + 7}}{{2n + 1}} = 4 + \dfrac{7}{{2n + 1}}\\
A \in N \Leftrightarrow \dfrac{7}{{2n + 1}} \in N\\
\Leftrightarrow 2n + 1 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
2n + 1 = 7\\
2n + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
n = 3\\
n = 0
\end{array} \right.\left( {TM} \right)\\
b)B = \dfrac{{6n + 13}}{{3n + 1}} = \dfrac{{2\left( {3n + 1} \right) + 11}}{{3n + 1}} = 2 + \dfrac{{11}}{{3n + 1}}\\
B \in N \Leftrightarrow \dfrac{{11}}{{3n + 1}} \in N\\
\Leftrightarrow 3n + 1 \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
3n + 1 = 11\\
3n + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
n = \dfrac{{10}}{3}\left( l \right)\\
n = 0\left( {TM} \right)
\end{array} \right.\\
c)C = \dfrac{{8n + 6}}{{4n + 2}} = \dfrac{{2\left( {4n + 2} \right) + 2}}{{4n + 2}} = 2 + \dfrac{2}{{4n + 2}}\\
C \in N \Leftrightarrow \dfrac{2}{{4n + 2}} \in N\\
\Leftrightarrow 4n + 2 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
4n + 2 = 2\\
4n + 2 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
n = 0\left( {TM} \right)\\
n = - \dfrac{1}{4}\left( l \right)
\end{array} \right.\\
d)D = \dfrac{{5n + 14}}{{n + 2}} = \dfrac{{5\left( {n + 2} \right) + 4}}{{n + 2}} = 5 + \dfrac{4}{{n + 2}}\\
D \in N \Leftrightarrow \dfrac{4}{{n + 2}} \in N\\
\Leftrightarrow n + 2 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
n + 2 = 4\\
n + 2 = 2\\
n + 2 = 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 2\\
n = 0\\
n = - 1\left( l \right)
\end{array} \right.\\
e)E = \dfrac{{6n + 13}}{{n + 2}} = \dfrac{{6\left( {n + 2} \right) + 1}}{{n + 2}} = 6 + \dfrac{1}{{n + 2}}\\
E \in N \Leftrightarrow \dfrac{1}{{n + 2}} \in N\\
\Leftrightarrow n + 2 \in U\left( 1 \right)\\
\to n + 2 = 1 \to n = - 1\left( l \right)\\
\end{array}\)
⇒ Vậy không có giá trị n thỏa mãn điều kiện
