Gọi d= UCLN(2n+3;3n+4)
\(\Rightarrow\left\{\begin{matrix}2n+3⋮d\\3n+4⋮d\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3\left(2n+3\right)\\2\left(3n+4\right)\end{matrix}\right.\Rightarrow\left\{\begin{matrix}6n+9\\6n+8\end{matrix}\right.\)
\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy: UCLN(2n+3; 3n+4) = 1