tìm n∈N 3n-1chia hết n+3 3n-1 chia hết 2-n 6n^2-1 chia hết 2n-3 n^2-7 chia hết n+2

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1. Đáp án

   $\begin{array}{l}
   a)3n – 1 = 3.\left( {n + 3} \right) – 10\\
   Do:3\left( {n + 3} \right) \vdots \left( {n + 3} \right)\\
    \Rightarrow 10 \vdots \left( {n + 3} \right)\\
    \Rightarrow \left( {n + 3} \right) \in \left\{ {5;10} \right\}\left( {do:n \ge 0} \right)\\
    \Rightarrow n \in \left\{ {2;7} \right\}\\
   b)3n – 1 = 3.\left( {n – 2} \right) + 6 – 1\\
    = 3\left( {n – 2} \right) + 5\\
   DO:3\left( {n – 2} \right) \vdots \left( {2 – n} \right)\\
    \Rightarrow 5 \vdots \left( {2 – n} \right)\\
    \Rightarrow \left( {2 – n} \right) \in \left\{ { – 5; – 1;1} \right\}\left( {do:2 – n \le 2} \right)\\
    \Rightarrow n \in \left\{ {7;3;1} \right\}\\
   c)6{n^2} – 1\\
    = 3n.\left( {2n – 3} \right) + 9n – 1\\
   Do:3n\left( {2n – 3} \right) \vdots \left( {2n – 3} \right)\\
    \Rightarrow 9n – 1 \vdots \left( {2n – 3} \right)\\
    \Rightarrow 2\left( {9n – 1} \right) \vdots \left( {2n – 3} \right)\\
    \Rightarrow 18n – 2 \vdots \left( {2n – 3} \right)\\
    \Rightarrow 9\left( {2n – 3} \right) + 25 \vdots \left( {2n – 3} \right)\\
    \Rightarrow \left( {2n – 3} \right) \in \left\{ { – 1;1;5;25} \right\}\left( {do:2n – 3 \ge  – 3} \right)\\
    \Rightarrow 2n \in \left\{ {2;4;8;28} \right\}\\
    \Rightarrow n \in \left\{ {1;2;4;14} \right\}\\
   d){n^2} – 7\\
    = n\left( {n + 2} \right) – 2n – 4 – 3\\
    = \left( {n + 2} \right).n – 2\left( {n + 2} \right) – 3\\
    \Rightarrow 3 \vdots \left( {n + 2} \right)\\
    \Rightarrow \left( {n + 2} \right) = 3\\
    \Rightarrow n = 1
   \end{array}$

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