\(A=\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)\)
\(=3n-2n^2-3+2n-\left(n^2+5n\right)\)
\(=3n-2n^2-3+2n-n^2-5n\)
\(=\left(3n-5n+2n\right)-\left(2n^2-n^2\right)-3\)
\(=-3\)
\(\Rightarrowđpcm\)
\(A=\left(n-1\right)\left(3-2n\right)-n\left(n+5\right) \)
\(=3n-2n^2-3+2n-\left(n^2+5n\right)\)
\(=3n-2n^2-3+2n-n^2-5n\)
\(=-3n^2-3\)
\(=3\left(-n^2-1\right)\)
Mà \(3\left(-n^2-1\right)⋮3\)
Vậy \(A⋮3\forall n\)