a, gọi d là ƯCLN(2n+1, 5n+2 )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\5n+2⋮d\end{cases}}\Rightarrow\hept{\begin{cases}5\left(2n+1\right)⋮d\\2\left(5n+2\right)⋮d\end{cases}}\Rightarrow\hept{\begin{cases}10n+5⋮d\\10+4⋮d\end{cases}}\)
\(\Rightarrow\left(10+5\right)-\left(10+4\right)⋮d\)
\(\Rightarrow10+5-10-4⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=\left\{-1;1\right\}\)
vậy...............