\(\left(6n+8\right)\)\(⋮\)\(\left(2n+1\right)\)
\(\Leftrightarrow\)\(3\left(2n+1\right)+5\)\(⋮\)\(2n+1\)
Ta thấy \(3\left(2n+1\right)\)\(⋮\)\(2n+1\)
nên \(5\)\(⋮\)\(2n+1\)
hay \(2n+1\)\(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta lập bảng sau:
\(2n+1\) \(-5\) \(-1\) \(1\) \(5\)
\(n\) \(-3\) \(-1\) \(0\) \(2\)
Vậy....