\(\left[ \begin{array}{l} n = 11\\ n = – 13\\ n = 5\\ n = – 7\\ n = 3\\ n = – 5\\ n = 2\\ n = – 4\\ n = 1\\ n = – 3\\ n = 0\\ n = – 2 \end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l} A = \dfrac{1}{3} + \dfrac{4}{{n + 1}}\\ = \dfrac{{n + 1 + 12}}{{3\left( {n + 1} \right)}} = \dfrac{{n + 13}}{{3\left( {n + 1} \right)}}\\ \to 3A = \dfrac{{n + 13}}{{n + 1}} = \dfrac{{n + 1 + 12}}{{n + 1}} = 1 + \dfrac{{12}}{{n + 1}}\\ A \in Z \Leftrightarrow \dfrac{{12}}{{n + 1}} \in Z\\ \to n + 1 \in U\left( {12} \right)\\ \to \left[ \begin{array}{l} n + 1 = 12\\ n + 1 = – 12\\ n + 1 = 6\\ n + 1 = – 6\\ n + 1 = 4\\ n + 1 = – 4\\ n + 1 = 3\\ n + 1 = – 3\\ n + 1 = 2\\ n + 1 = – 2\\ n + 1 = 1\\ n + 1 = – 1 \end{array} \right. \to \left[ \begin{array}{l} n = 11\\ n = – 13\\ n = 5\\ n = – 7\\ n = 3\\ n = – 5\\ n = 2\\ n = – 4\\ n = 1\\ n = – 3\\ n = 0\\ n = – 2 \end{array} \right. \end{array}\)
Đáp án:
`n∈{0;-2;1;-3;2;-4;3;-5;5;-7;11;-13}`
Giải thích các bước giải:
`A=1/3+4/(n+1)`
`->A=(n+1)/(3(n+1))+12/(3(n+1))`
`->A=(n+13)/(3(n+1))`
`->3A=(n+13)/(n+1)=1+12/(n+1)`
`A∈Z→12/(n+1)∈Z→n+1∈Ư(12)={±1;±2;±3;±4;±6;±12}`
Ta có bảng
\begin{array}{|c|c|}\hline n+1&1&-1&2&-2&3&-3&4&-4&6&-6&12&-12\\\hline n&0&-2&1&-3&2&-4&3&-5&5&-7&11&-13\\\hline \end{array}
Vậy `n∈{0;-2;1;-3;2;-4;3;-5;5;-7;11;-13}`
Đáp án:
\(\left[ \begin{array}{l}
n = 11\\
n = – 13\\
n = 5\\
n = – 7\\
n = 3\\
n = – 5\\
n = 2\\
n = – 4\\
n = 1\\
n = – 3\\
n = 0\\
n = – 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{1}{3} + \dfrac{4}{{n + 1}}\\
= \dfrac{{n + 1 + 12}}{{3\left( {n + 1} \right)}} = \dfrac{{n + 13}}{{3\left( {n + 1} \right)}}\\
\to 3A = \dfrac{{n + 13}}{{n + 1}} = \dfrac{{n + 1 + 12}}{{n + 1}} = 1 + \dfrac{{12}}{{n + 1}}\\
A \in Z \Leftrightarrow \dfrac{{12}}{{n + 1}} \in Z\\
\to n + 1 \in U\left( {12} \right)\\
\to \left[ \begin{array}{l}
n + 1 = 12\\
n + 1 = – 12\\
n + 1 = 6\\
n + 1 = – 6\\
n + 1 = 4\\
n + 1 = – 4\\
n + 1 = 3\\
n + 1 = – 3\\
n + 1 = 2\\
n + 1 = – 2\\
n + 1 = 1\\
n + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 11\\
n = – 13\\
n = 5\\
n = – 7\\
n = 3\\
n = – 5\\
n = 2\\
n = – 4\\
n = 1\\
n = – 3\\
n = 0\\
n = – 2
\end{array} \right.
\end{array}\)