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

hay \(n\in\left\{3;1;7;-3\right\}\)

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

\(\dfrac{1}{2}.2^{n+4}.2^n=2^5\\ =>2^{n+4+n}=2^5:\dfrac{1}{2}\\ =>2^{2n+4}=2^5.2\\ =>2^{2n+4}=2^6\\ =>2n+4=6\\ =>2n=2=>n=1\)

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{1}{2}\cdot2^{n+4}\cdot2^n=2^5\)

`\Rightarrow `\(\dfrac{1}{2}\cdot2^n\cdot2^4\cdot2^n=2^5\)

`\Rightarrow `\(2^{n\cdot2+4}=2^5\div\dfrac{1}{2}\)

`\Rightarrow `\(2^{n\cdot2+4}=2^6\)

`\Rightarrow `\(n\cdot2+4=6\)

`\Rightarrow `\(2n=2\)

`\Rightarrow n=1`

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

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

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}

a: \(A=\left(1+3\right)+...+3^{10}\left(1+3\right)\)

\(=4\left(1+...+3^{10}\right)⋮4\)