Question

Tìm số tự nhiên n sao cho 4n + 7 chia hết cho 2n + 1

Answer 1

=>4n+2+5 chia hết cho 2n+1

=>5 chia hết cho 2n+1

\(\Leftrightarrow2n+1\in\left\{1;5;-1;-5\right\}\)

hay \(n\in\left\{0;2\right\}\)

Answer 2

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}