Gọi d là ƯCLN (2n + 3 ; 3n + 4)
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\3n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+3\right)⋮d\\2\left(3n+4\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}6n+9⋮d\\6n+8⋮d\end{cases}}}\)
\(\Rightarrow6n+9-\left(6n+8\right)⋮d\)
\(\Rightarrow6n+9-6n-8⋮d\)
\(\Rightarrow9-8⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ƯCLN (2n + 3 ; 3n + 4) = 1