\(\left(2n+3\right)^2-\left(2n-1\right)^2=4n^2+12n+9-4n^2+4n-1=16n+8=8\left(2n+1\right)⋮8\)
\(\left(2n+3\right)^2-\left(2n-1\right)^2\)
\(=\left(2n+3-2n+1\right)\left(2n+3+2n-1\right)\)
\(=4\left(4n-2\right)\)
\(=8\left(2x-1\right)\) Vì \(8⋮8\)
\(\Rightarrow8\left(2n-1\right)⋮(ĐPCM)\)