Bài 5:
Ta có:
\(Q=\left(2n-1\right)\left(2n+3\right)-\left(4n-5\right)\left(n+1\right)+3\)
\(Q=4n^2+6n-2n-3-\left(4n^2+4n-5n-5\right)+3\)
\(Q=4n^2+4n-3+3-4n^2+n+5\)
\(Q=\left(4n^2-4n^2\right)+\left(4n+n\right)+\left(-3+3+5\right)\)
\(Q=5n+5\)
\(Q=5\left(n+1\right)\)
Mà: \(5\left(n+1\right)\) ⋮ 5 ∀ n
\(\Rightarrow Q\) ⋮ 5 ∀ n



