Question

Tìm n ϵ N: 4n+3 chia hết cho 2n + 1

Answer 1

Vì : \(2n+1⋮2n+1\Rightarrow2\left(2n+1\right)⋮2n+1\Rightarrow4n+2⋮2n+1\)

Mà : \(4n+3⋮2n+1\)

\(\Rightarrow\left(4n+3\right)-\left(4n+2\right)⋮2n+1\)

\(\Rightarrow4n+3-4n-2⋮2n+1\)

\(\Rightarrow1⋮2n+1\Rightarrow2n+1=1\Rightarrow n=0\)

Vậy n = 0 thỏa mãn

Answer 2

ta có:

4n+3\(⋮\)2n+1

4n+2+1\(⋮\)`2n+1

2(2n+1)+1\(⋮\)2n+1

Vì 2(n+1)\(⋮\)2n+1 nên 1\(⋮\)2n+1

=>2n+1 là Ư(1)

Ư(1)={1;-1}

n={0;-1}