a) Ta có: \(\hept{\begin{cases}2n+1⋮6-n\\6-n⋮6-n\end{cases}\Rightarrow\hept{\begin{cases}2n+1⋮6-n\\2.\left(6-n\right)⋮6-n\Rightarrow12-2n⋮6-n\end{cases}\Rightarrow}2n+1+12-2n⋮6-n}\)
\(\Rightarrow13⋮6-n\Rightarrow6-n\inƯ\left(13\right)=\left\{1;13\right\}\Rightarrow n\in\left\{5;-7\right\}\). Mà \(n\in N\Rightarrow n=5\)
b) \(\hept{\begin{cases}3n⋮n-1\\3\left(n-1\right)⋮n-1\end{cases}\Rightarrow\hept{\begin{cases}3n⋮n-1\\3n-3⋮n-1\end{cases}\Rightarrow}3n-\left(3n-3\right)⋮n-1}\)
\(\Rightarrow3n-3n+3⋮n-1\Rightarrow3⋮n\Rightarrow n\inƯ\left(3\right)\). Mà \(n\in N\Rightarrow n\in\left\{1;3\right\}\)
c) \(\hept{\begin{cases}3n+5⋮2n+1\\2n+1⋮2n+1\end{cases}\Rightarrow\hept{\begin{cases}2.\left(3n+5\right)⋮2n+1\\3.\left(2n+1\right)⋮2n+1\end{cases}\Rightarrow}\hept{\begin{cases}6n+10⋮2n+1\\6n+3⋮2n+1\end{cases}}}\)
\(\Rightarrow6n+10-\left(6n+3\right)⋮2n+1\Rightarrow6n+10-6n-3⋮2n+1\Rightarrow7⋮2n+1\)
\(\Rightarrow2n+1\inƯ\left(7\right)=\left\{1;7;-1;-7\right\}\Rightarrow2n\in\left\{0;6;-2;-8\right\}\Rightarrow n\in\left\{0;3;-1;-4\right\}\)
Mà \(n\in N\Rightarrow n\in\left\{0;3\right\}\)
