Đặt 3 . (n + 2) \(⋮\) n - 2
\(\Rightarrow3.2+n\) chia hết cho \(n-2\)
\(\Rightarrow6+n\) chia hết cho \(n-2\)
\(\Rightarrow n=6:2=3\)
\(3\left(n+2\right)=3n+6=3n-6+12=3\left(n-2\right)+12\)
\(3\left(n-2\right)⋮n-2\)
Để \(3\left(n+2\right)⋮n-2\) thì \(12⋮n-2\Rightarrow n-2\inƯ\left(12\right)\)
\(Ư\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
\(\Rightarrow n-2\in\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
\(\Rightarrow n\in\left\{-10;-4;-2;-1;0;1;3;4;5;6;8;14\right\}\)
3(n + 2) \(⋮n-2\)
<=> 3(n - 2) + 12 \(⋮n-2\)
<=> 12 \(⋮n-2\)
=> n - 2 \(\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
=> n = 3; 1; 4; 0; 5; -1; 6; -2; 8; -4; 14; -10
@Thảo Nguyễn