\( \to \left[ \begin{array}{l} n – 4 = 1\\ n – 4 = – 1\\ n – 4 = 2\\ n – 4 = – 2\\ n – 4 = 3\\ n – 4 = – 3\\ n – 4 = 6\\ n – 4 = – 6\\ n – 4 = 9\\ n – 4 = – 9\\ n – 4 = 18\\ n – 4 = – 18 \end{array} \right. \leftrightarrow \left[ \begin{array}{l} n = 5\\ n = 3\\ n = 6\\ n = 2\\ n = 7\\ n = 1\\ n = 10\\ n = – 2\\ n = 13\\ n = – 5\\ n = 22\\ n = – 14 \end{array} \right.\)
mà n∈N
\( \to \left[ \begin{array}{l} n = 5\\ n = 3\\ n = 6\\ n = 2\\ n = 7\\ n = 1\\ n = 10\\ n = 13\\ n = 22 \end{array} \right.\)
Đáp án:
\(\left[ \begin{array}{l}
n = 5\\
n = 3\\
n = 6\\
n = 2\\
n = 7\\
n = 1\\
n = 10\\
n = 13\\
n = 22
\end{array} \right.\)
Giải thích các bước giải:
\(\frac{{{n^2} + 2n – 6}}{{n – 4}} = \frac{{{n^2} – 4n + 6n – 24 + 18}}{{n – 4}} = \frac{{n(n – 4) + 6(n – 4) + 18}}{{n – 4}} = n + 6 + \frac{{18}}{{n – 4}}\)
Để (n² +2n -6) chia hết cho (n-4)
<-> 18 chia hết cho n-4
\( \to \left[ \begin{array}{l}
n – 4 = 1\\
n – 4 = – 1\\
n – 4 = 2\\
n – 4 = – 2\\
n – 4 = 3\\
n – 4 = – 3\\
n – 4 = 6\\
n – 4 = – 6\\
n – 4 = 9\\
n – 4 = – 9\\
n – 4 = 18\\
n – 4 = – 18
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
n = 5\\
n = 3\\
n = 6\\
n = 2\\
n = 7\\
n = 1\\
n = 10\\
n = – 2\\
n = 13\\
n = – 5\\
n = 22\\
n = – 14
\end{array} \right.\)
mà n∈N
\( \to \left[ \begin{array}{l}
n = 5\\
n = 3\\
n = 6\\
n = 2\\
n = 7\\
n = 1\\
n = 10\\
n = 13\\
n = 22
\end{array} \right.\)
Ta có: $n^2+2n-6$ ⋮ $n-4$
$⇒(n^2-4n)+(6n-24)+18$ ⋮ $n-4$
$⇒18$ ⋮ $n-4$
$⇒n-4∈Ư(18)=${$±1;±2;±3;±6;±9;±18$}
Vì $n∈N⇒n≥0$ nên $n-4≥-4⇒n-4∈${$±1;±2;±3;6;9;18$}
Lập bảng:
n-4 -1 1 -2 2 -3 3 6 9 18
n 3 5 2 6 1 7 10 13 22
Vậy $n∈${$3;5;2;6;1;7;10;13;22$}.