\(\left(5n+12\right)⋮\left(n-3\right)\)

\(\Rightarrow5\left(n-3\right)+27⋮\left(n-3\right)\)

\(\Rightarrow\left(n-3\right)\inƯ\left(27\right)=\left\{1;-1;3;-3;9;-9;27;-27\right\}\)

Do \(n\in N\)

\(\Rightarrow n\in\left\{0;2;4;6;12;30\right\}\)

Ta có: 5n+12 ⋮ n-3

      ⇔ 5(n-3)+27 ⋮ n-3

      ⇔ 27⋮n-3 hay n-3∈Ư_(27​)

mà n là số tự nhiên

⇒ n-3≥-3

hay n-3={-3;-1;1;3;9;27}

  ⇔ n= {0;2;4;6;12;30}

Ta có: \(5n+12⋮n-3\)

\(\Leftrightarrow5n-15+27⋮n-3\)

\(\Leftrightarrow n-3\in\left\{1;-1;3;-3;9;-9;27;-27\right\}\)

hay \(n\in\left\{4;2;6;0;12;-6;30;-24\right\}\)

\(3^{5n+2}+3^{5n+1}-3^{5n}=3^{5n}\left(3^2+3-1\right)=11.3^{5n}⋮11\)

\(3^{5n+2}+3^{5n+1}-3^{5n}(n\in N^*)\\=3^{5n}\cdot3^2+3^{5n}\cdot3-3^{5n}\\=3^{5n}\cdot(3^2+3-1)\\=3^{5n}\cdot11\)

Vì \(3^{5n}\cdot11\vdots11\) 

nên biểu thức \(3^{5n+2}+3^{5n+1}-3^{5n}\vdots11\)

(3n-5)(2n+1)+7(n-1)=6n²-7n-5+7n-7

                           =6n²-12

                           =3(2n-4)

=>(3n-5)(2n+1)+7(n-1) chia hết cho 3, với mọi n

(n-4)(5n+3)-(n+1)(5n-2)+4=5n²-17n-12-(5n²+3n-2)

 =5n²-17n-12-5n²-3n+2

=-20n-10

=5(-4n-2)

=>(n-4)(5n+3)-(n+1)(5n-2)+4 chia hết cho 5, với mọi n

trieu dang làm đúng rùi