\(a,\Rightarrow n-2+5⋮n-2\\ \Rightarrow n-2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow n\in\left\{-3;1;3;7\right\}\\ b,\Rightarrow2\left(n-4\right)+13⋮n-4\\ \Rightarrow n-4\inƯ\left(13\right)=\left\{-13;-1;1;13\right\}\\ \Rightarrow n\in\left\{-9;3;5;17\right\}\\ c,\Rightarrow6n-9⋮3n+1\\ \Rightarrow2\left(3n+1\right)-12⋮3n+1\\ \Rightarrow3n+1\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\\ \Rightarrow n\in\left\{-1;0;1\right\}\left(n\in Z\right)\\ d,\Rightarrow n^2+2n-n-2+3⋮n+2\\ \Rightarrow n\left(n+2\right)-\left(n+2\right)+3⋮n+2\\ \Rightarrow n+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Rightarrow n\in\left\{-5;-3;-1;1\right\}\)

a: \(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{3;1;7;-3\right\}\)

https://loga.vn/hoi-dap/tim-so-tu-nhien-n-sao-cho-n-6-chia-het-cho-n-2-36137

https://h7.net/hoi-dap/toan-6/tim-so-tu-nhien-n-de-3n-7-chia-het-cho-n-faq26687.html

n(2n-3)-2n(n+2)

=2n²-3n-2n²-4n

= - 7n luôn chia hết cho 7 (vì -7 chia hết cho 7)

vậy n(2n-3)-2n(n+2) luôn chia hết cho 7 với mọi n

tham khảo ở link bn nhé

a) Theo bài ra: 4n - 7 ⋮ n - 1

⇔ \(\frac{4n-7}{n-1}\in Z\)

Suy ra, ta có: \(\frac{4n-7}{n-1}=\frac{n-1+n-1+n-1+n-1-3}{n-1}=\frac{n-1}{n-1}+\frac{n-1}{n-1}+\frac{n-1}{n-1}+\frac{n-1}{n-1}-\frac{3}{n-1}\)

\(=1+1+1+1-\frac{3}{n-1}=4-\frac{3}{n-1}\)

Để \(4-\frac{3}{n-1}\in Z\) thì \(\frac{-3}{n-1}\in Z\)

\(\Rightarrow n-1\inƯ\left(3\right)\)

Mà Ư(3) = {\(\pm1\pm3\)}

Do đó: n - 1 = 1 ⇔ n = 2

hoặc n - 1 = -1 ⇔ n = 0

hoặc n - 1 = 3 ⇔ n = 4

hoặc n - 1 = -3 ⇔ n - 1 = -2 (loại vì -2∉ N)

Vậy n = {2;0;4)

Câu b làm tương tự nhé!

\(n+1;n;n-1\)

n+1;n;n−1

A

\(24+n⋮n\Rightarrow24⋮n\)

\(\Rightarrow n\in\left\{1;2;3;4;6;8;12;24\right\}\)

b) \(15-n⋮n\Rightarrow15⋮n\)

\(\Rightarrow n\in\left\{1;3;5;15\right\}\)

\(n+15⋮n+3\)

\(\Rightarrow12⋮n+3\)

\(\Rightarrow n+3\in\left\{3;4;6;12\right\}\)

\(\Rightarrow n\in\left\{0;1;3;9\right\}\)