\(\Rightarrow3\left(2n+3\right)⋮3n+1\\ \Rightarrow6n+9⋮3n+1\\ \Rightarrow2\left(3n+1\right)+7⋮3n+1\\ \Rightarrow2n+1\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow n\in\left\{0;3\right\}\)
Ta có: \(\left(2n+3\right)⋮\left(3n+1\right)\)
\(\Rightarrow3\left(2n+3\right)⋮\left(3n+1\right)\)
\(\Rightarrow6n+9⋮3n+1\)
\(\Rightarrow3n+1+3n+1+7⋮3n+1\)
Do \(3n+1⋮3n+1\)
\(\Rightarrow7⋮3n+1\Rightarrow3n+1\leftarrowƯ\left(7\right)=\left\{1;7\right\}\)
Với \(3n+1=1\Rightarrow n=0\)
\(3n+1=7\Rightarrow n=2\)
Vậy \(n\in\left\{0;2\right\}\)