\(a,\text{Gọi }d=ƯCLN\left(n,n+1\right)\\ \Rightarrow n⋮d;n+1⋮d\\ \Rightarrow n+1-n⋮d\\ \Rightarrow1⋮d\Rightarrow d=1\RightarrowƯCLN\left(n,n+1\right)=1\\ \RightarrowĐpcm\\ b,\text{Gọi }d=ƯCLN\left(2n+1,2n\right)\\ \Rightarrow2n+1⋮d;2n⋮d\\ \Rightarrow2n+1-2n=1⋮d\\ \Rightarrow d=1\\ \RightarrowƯCLN\left(2n+1,2n\right)=1\\ \RightarrowĐpcm\\ c,\text{Gọi }d=ƯCLN\left(n+1,2n+3\right)\\ \Rightarrow n+1⋮d;2n+3⋮d\\ \Rightarrow2n+3-2\left(n+1\right)=1⋮d\\ \Rightarrow d=1\\ \RightarrowƯCLN\left(n+1,2n+3\right)=1\\ \RightarrowĐpcm\)