\(\left(2n+3\right)^2-9=\left(2n+3\right)^2-3^2=\left(2n+3-3\right)\left(2n+3+3\right)=2n\left(2n+6\right)=4n\left(n+3\right)⋮4\)
\(\left(3n+4\right)^2-16=\left(3n+4\right)^2-4^2=\left(3n+4-4\right)\left(3n+4+4\right)=3n\left(3n+8\right)⋮3\)
a) ta có: (2n+3)2 - 9
= 4n2 +12n + 9 - 9
= 4n.(n+3) chia hết cho 4
=> ...
b) ta có: (3n+4)2 - 16
= 9n2 + 24n + 16 - 16
= 3n.(3n + 8) chia hết cho 3
=> ...