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

hay \(n\in\left\{3;1;7;-3\right\}\)

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

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

3n + 4 ⋮ n + 1

=> 3n + 3 + 1 ⋮ n + 1

=> 3(n + 1) + 1  ⋮ n + 1

=> 1 ⋮ n + 1

=> n + 1 thuộc Ư(1)

=> n + 1 thuộc {-1;1}

=> n thuộc {-2;0}

=>3n+3+1:n+1

=>1:n+1

=>n+1 E Ư(1)={1;-1}

=>n={0;-2}

Vậy n={0;-2}

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

n+1;n;n−1

A

a: Để B là số nguyên thì \(n+1⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{3;1;5;-1\right\}\)

b: \(B=\dfrac{n-2+3}{n-2}=1+\dfrac{3}{n-2}\)

Để B có giá trị lớn nhất thì n-2=-1

hay n=1