3n+2 ⋮ 3n+1
3n+1-1+2 ⋮ 3n+1
3n+1+1 ⋮ 3n+1
Vì 3n+1 ⋮ 3n+1 nên 1⋮ 3n+1
⇒3n+1 ∈ Ư(1)
⇒ 3n+1 ∈ ξ 1 ξ
⇒ 3n ∈ ξ 0 ξ
⇒ n ∈ ξ 0ξ
Vậy n=0
\(3n+2⋮3n+1.\)
\(\Rightarrow\left(3n+1\right)+1⋮3n+1.\)
mà \(3n+1⋮3n+1\Rightarrow1⋮3n+1\in U_{\left(1\right)}=\left\{\pm1\right\}.\)
\(\Rightarrow n\in\left\{0;-\dfrac{2}{3}\right\}.\)
Vậy..........