Đặt \(d=\left(2n+1,2n^2-1\right)\).
\(\hept{\begin{cases}2n+1⋮d\\2n^2-1⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2n^2+n⋮d\\2n^2-1⋮d\end{cases}\Rightarrow}\left[\left(2n^2+n\right)-\left(2n^2-1\right)\right]⋮d\)
\(\Rightarrow\left(n+1\right)⋮d\Rightarrow\left[2\left(n+1\right)-\left(2n+1\right)\right]⋮d\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\Rightarrow\left(2n+1,2n^2-1\right)=1\)
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