Đáp án+Giải thích các bước giải:
`C1:` Ta có:
`(2n+3)^2-9`
`=(2n)^2+2.2n.3+3^2-9`
`=4n^2+12n+9-9`
`=4n^2+12n`
`=4(n^2+3n)`
Vì `4\vdots4`
`=>` `4(n^2+3n)\vdots4` `∀` `n`
`=>` `(2n+3)^2-9\vdots4` `∀` `n`
Vậy `(2n+3)^2-9\vdots4` `∀` `n`
`C2:` Ta có:
`(2n+3)^2-9`
`=(2n+3)^2-3^2`
`=(2n+3-3)(2n+3+3)`
`=2n(2n+6)`
`=2n.2(n+3)`
`=4n(n+3)`
Vì `4\vdots4`
`=>` `4n\vdots4` `∀` `n`
`=>` `4n(n+3)\vdots4` `∀` `n`
`=>(2n+3)^2-9\vdots4` `∀` `n`
Vậy `(2n+3)^2-9\vdots4` `∀` `n`
