a) \(4\left(n-1\right)-3⋮\left(n-1\right)\)

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

Do \(n\in N\Rightarrow n\in\left\{0;2;4\right\}\)

b) \(-5\left(4-n\right)+12⋮\left(4-n\right)\)

\(\Rightarrow\left(4-n\right)\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)

Do \(n\in N\Rightarrow n\in\left\{16;10;8;7;6;5;3;2;1;0\right\}\)

c) \(-2\left(n-2\right)+6⋮\left(n-2\right)\)

\(\Rightarrow\left(n-2\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Do \(n\in N\Rightarrow n\in\left\{0;1;3;4;5;8\right\}\)

d) \(n\left(n+3\right)+6⋮\left(n+3\right)\)

\(\Rightarrow\left(n+3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Do \(n\in N\Rightarrow n\in\left\{0;3\right\}\)

1. a) \(\left(n+15\right)⋮\left(n+2\right)\)

\(\Rightarrow\left[n+15-\left(n+2\right)\right]⋮\left(n+2\right)\)

\(\Rightarrow\left[n+15-n-2\right]⋮\left(n+2\right)\)

\(\Rightarrow13⋮\left(n+2\right)\)

\(\Rightarrow\left(n+2\right)\inƯ_{\left(13\right)}=\left\{\pm1;\pm13\right\}\)

\(\Rightarrow n\in\left\{...\right\}\)

b) \(\left(3n+17\right)⋮\left(n+1\right)\)

\(\Rightarrow\left(3n+17\right)⋮3\left(n+1\right)\)

\(\Rightarrow\left(3n+17\right)⋮\left(3n+3\right)\)

\(\Rightarrow\left[\left(3n+17\right)-\left(3n+3\right)\right]⋮\left(n+1\right)\)

\(\Rightarrow\left[3n+17-3n-3\right]⋮\left(n+1\right)\)

\(\Rightarrow14⋮\left(n+1\right)\)

\(\Rightarrow\left(n+1\right)\inƯ_{\left(14\right)}=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

\(\Rightarrow n\in\left\{...\right\}\)

a, \(n+15⋮n+2\)

\(\Rightarrow n+2+13⋮n+2\)

\(\Rightarrow n+2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

+ n + 2 = 1 \(\Rightarrow\)n = -1

+ n + 2 = -1 \(\Rightarrow\)n = -3

+ n + 2 = -13 \(\Rightarrow\)n = -15

+ n + 2 = 13 \(\Rightarrow\)n = 11

b, \(3n+17⋮n+1\)

\(\Rightarrow3\left(n+1\right)+14⋮n+1\)

\(\Rightarrow n+1\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

+ n + 1 = 1\(\Rightarrow\)n = 0

+ n + 1 = -1\(\Rightarrow\)n = -2

+ n + 1 = 2\(\Rightarrow\)n = 1

+ n + 1 = -2\(\Rightarrow\)n = -3

+ n + 1 = 7\(\Rightarrow\)n = 6

+ n + 1 = -7\(\Rightarrow\)n = -8

+ n + 1 = 14\(\Rightarrow\)n = 13

+ n + 1 = -14\(\Rightarrow\)n = -15

a/n+2 chia hết cho n-1

=>(n-1)+3 chia hết cho n-1

=>n-1 thuộc Ư(3)={1;3}

n-1=1=>n=2

n-1=3=>n=4

=>n E {2;4}

b/

2n+1 chia hết chon+ 1

=>2(n+1)-1 chia hết cho n+1

=>1 chia hết cho n+1

=>n+1=1

=>n=0

a) n+3 chia hết cho n-2

=>n-2+5 chia hết cho n-2

=> 5 chia hết cho n-2

U(5)=1;5

=>n=3;7 

Ta có: n + 3 chia hết cho n - 2

<=> n - 2 + 5 chia hết n - 2

=> 5 chia hết n - 2

=> n - 2 thuộc Ư(5) = {-1;1;-5;5}

=> n = {1;3;-3;7}

b)\(\frac{2n+5}{n+1}=\frac{2\left(n+1\right)+3}{n+1}=\frac{2\left(n+1\right)}{n+1}+\frac{3}{n+1}=2+\frac{3}{n+1}\in Z\)

=>3 chia hết n+1

=>n+1 thuộc Ư(3)={1;3} (vì n thuộc N)

=>n thuộc {0;2}

c)\(\frac{4n+3}{2n+6}=\frac{2\left(2n+6\right)-9}{2n+6}=\frac{2\left(2n+6\right)}{2n+6}-\frac{9}{2n+6}=2-\frac{9}{2n+6}\in Z\)

=>9 chia hết 2n+6

=>2n+6 thuộc Ư(9)={1;3;9} (vì n thuộc N)

=>n thuộc rỗng 

n+5 chia hết cho n+2

=> n+2+3 chia chia hết cho n+2

mà n+2 chia hết n+2

=>3 chia hết cho n+2

n+2  -3; -1; 1; 3

n     -5; -3; -1; 1

Vậy tập hợp các số n thỏa mãn là A={-5;-3;-1;1}

\(\frac{n+5}{n+2}=\frac{n+2+3}{n+2}=1+\frac{3}{n+2}\)

Vì 1 là số tự nhiên nên để n+5\(⋮\)n+3 thì 3\(⋮\)n+2.

Vậy (n+2)\(\in\)Ư(3)=>n+2\(\in\){-3;-1;1;3}

=>n\(\in\){-5;-3;-1;1}

Mà n \(\in\)N nên n = 1.

\(n+5⋮n+2\)

\(=n+2+3⋮n+2\)

\(\Rightarrow3⋮n+2\)

\(n+2\inư\left(3\right)\in1,3\)

Vậy n = 1

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

hay \(n\in\left\{3;1;7\right\}\)