Gọi ƯCLN(4n+3,5n+1)=d(d\(\inℕ^∗\))
\(\Rightarrow\)4n+3\(⋮\)d
5n+1\(⋮\)d
\(\Rightarrow\)5.(4n+3)\(⋮\)d
4.(5n+1)\(⋮\)d
\(\Rightarrow\)20n+15\(⋮\)d
20n+4\(⋮\)d
\(\Rightarrow\)(20n+15-20n-4)\(⋮\)d
\(\Rightarrow\)11\(⋮\)d
Do đó d \(\in\)Ư(11)={1;11}
Mà đầu bài cho là (4n+3,5n+1)\(\ne\)1
\(\Rightarrow\)d=11
Vậy ƯCLN(4n+3,5n+1)=11