Question

(4n+3) chia hết cho (2n+1)

Answer 1

Ta có : \(\frac{4n+3}{2n+1}\Rightarrow\frac{4n+2+1}{2n+1}\Rightarrow\frac{2\left(2n+1\right)+1}{2n+1}\Rightarrow\frac{1}{2n+1}\)

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

Vậy n = 0

Answer 2

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

\(8n+4-1⋮2n+1\)

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

\(V\text{ì}2n+1⋮2n+1\)=>4(2n+1)\(⋮\)(2n+1)

Buộc 1 \(⋮\)(2n+1)=>2n+1ϵƯ(1)={1}

Với 2n+1=1=>2n=0=>n=0

Vậy n=0

2n+1	1	2	4
2n	2	\(\frac{3}{2}\)loại
n	1

Answer 3

4n+3=4n+4-1=4x[n+1]-1=2x[2n+1]-1

vì 2x[2n+1] chia hết cho 2n+1

\(\Rightarrow\)Muốn 2x[2n+1]-1 chia hết cho 2n+1 thì 1 chia hết cho 2n+1

\(\Rightarrow\)2n+1 \(\in\)Ư[1]={1}

*Lập bảng

2n+1 1 2n 0 n 0

Vậy n=0