=>4n+2+5 chia hết cho 2n+1
=>5 chia hết cho 2n+1
\(\Leftrightarrow2n+1\in\left\{1;5;-1;-5\right\}\)
hay \(n\in\left\{0;2\right\}\)
4n+7⋮2n+14n+7⋮2n+1
⇒4n+7−[2(2n+1)]⋮2n+1⇒4n+7−[2(2n+1)]⋮2n+1
⇒4n+7−(4n+2)⋮2n+1⇒4n+7−(4n+2)⋮2n+1
⇒4n+7−4n−2⋮2n+1⇒4n+7−4n−2⋮2n+1
⇒5⋮2n+1⇒5⋮2n+1
⇒2n+1∈Ư(5)={1;5}⇒2n+1∈Ư(5)={1;5}
⇒2n∈{0;4}⇒2n∈{0;4}
⇒n∈{0;2}⇒n∈{0;2}
Vậy n∈{0;2}