\(\left(4n+3\right)^2-25\)
\(=\left(4n+3\right)^2-5^2\)
\(=\left(4n+3-5\right)\left(4n+3+5\right)\)
\(=\left(4n-2\right)\left(4n+8\right)\)
Ta có ; \(\left(4n+3\right)^2-25=\left(4n+3\right)^2-5^2=\left(4n+3-5\right)\left(4n+3+5\right)\)
\(=\left(4n-2\right)\left(4n+8\right)=8\left(2n-1\right)\left(n+2\right)\)chia hết cho 8 với mọi số nguyên n