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

a) Vì 5n + 7 chia hết cho n

\(\Rightarrow7⋮n\Rightarrow n\inƯ\left(7\right)\Rightarrow n\in\left\{\pm1;\pm7\right\}\)

Vậy \(n\in\left\{\pm1;\pm7\right\}\)

b) Vì n + 9 chia hết cho n +4

\(\Rightarrow\left(n+4\right)+5⋮n+4\)

\(\Rightarrow5⋮n+4\)

\(\Rightarrow n+4\inƯ\left(5\right)\)

\(\Rightarrow n+4\in\left\{\pm1;\pm5\right\}\)

\(\Rightarrow n\in\left\{-3;-5;-1;-9\right\}\) \(\inℕ\)

Vậy \(n\in\left\{-3;-5;-1;-9\right\}\)

c, Vì 2n + 1 chia hết cho n - 3

\(\Rightarrow\left(2n-6+7\right)⋮\left(n-3\right)\)

\(\Rightarrow2\left(n-3\right)+7⋮n-3\)

\(2\left(n-3\right)⋮n-3\Rightarrow7⋮n-3\)

Phần còn lại lm như trên

Vì 35 chia hết cho 7

   77 chia hết cho 7

   6 không chia hết cho 7

Để A không chia hết cho 7 thì n phải chia hết cho 7

=> n thuộc { 7 ; 14 ; 28 ; 42 ; ... }

A, n=1

B, n khác 1

khác 1+ 7

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) Để n + 1 là ước của 2n + 7 thì :

2n + 7 ⋮ n + 1

2n + 2 + 5 ⋮ n + 1

2( n + 1 ) + 5 ⋮ n + 1

Vì 2( n +1 ) ⋮ n + 1

=> 5 ⋮ n + 1

=> n + 1 thuộc Ư(5) = { 1; 5; -1; -5 }

=> n thuộc { 0; 4; -2; -6 }

Vậy........ 

\(\text{n + 1 là ước của 2n + 7 nên }\left(2n+7\right)⋮\left(n+1\right)\)

\(\Rightarrow\left(2n+2+5\right)⋮\left(n+1\right)\)

\(\Rightarrow5⋮\left(n+1\right)\left[\text{vì }\left(2n+2\right)⋮\left(n+1\right)\right]\)

\(\Rightarrow n+1\inƯ\left(5\right)=\left\{1;5\right\}\)

\(\text{Trường hợp : }n+1=1\)

\(\Rightarrow n=1-1\)

\(\Rightarrow n=0\)

\(\text{Trường hợp : }n+1=5\)

\(\Rightarrow n=5-1\)

\(\Rightarrow n=4\)

\(\text{Vậy }n\in\left\{0;4\right\}\)