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
\(\left(4n+3\right)⋮\left(2n+1\right)\)
\(8n+4-1⋮2n+1\)
\(4\left(2n+1\right)-1⋮\left(2n+1\right)\)
\(V\text{ì}2n+1⋮2n+1\)=>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 |
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
|
Vậy n=0
