Question

Tìm UC(2n+1;3n+1) Với n thuộc N

Answer 1

Gọi UCLN(2n+1;3n+1) là d

Ta có:

[3(2n+1)]-[2(3n+1)] chia hết d

=>[6n+3]-[6n+2] chia hết d

=>1 chia hết d

=>d=1

Vậy UC(2n+1;3n+1)=1

Answer 2

\(G\text{ọi}dl\text{à}UCLN\left(2n+1;3n+1\right)\\ =>2n+1v\text{à}3n+1⋮d\\ =>\left(2n+1\right)-\left(3n+1\right)⋮d\\ =>3\left(2n+1\right)-\left(2\left(3n+1\right)\right)⋮d\)

\(=>6n+3-6n-2⋮d\\ =1⋮d\\ =>d=1\)

Vậy UCLN(2n+1;3n+1) là 1 hay UC (2n+1;3n+1) là 1