Ta có :
\(\left\{{}\begin{matrix}3n+4⋮2n+1\\2n+1⋮2n+1\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}2\left(3n+4\right)⋮2n+1\\3\left(2n+1\right)⋮2n+1\end{matrix}\right.\\ \rightarrow2\left(3n+4\right)-3\left(2n+1\right)⋮2n+1\\ \rightarrow5⋮2n+1\\ \rightarrow\left\{{}\begin{matrix}2n+1\inƯ\left(5\right)\\2n+1\in N\end{matrix}\right.\\ \rightarrow2n+1\in\left\{1;5\right\}\)
Vậy `n = 0` hoặc `n=2`
=>6n+8 chia hết cho 2n+1
=>6n+3+5 chia hết cho 2n+1
mà n là số tự nhiên
nên \(2n+1\in\left\{1;5\right\}\)
=>\(n\in\left\{0;2\right\}\)