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

\(n+3⋮n\cdot n-7\)

\(\Rightarrow n+3⋮n^2-7\)

\(\Rightarrow(n+3)(n+3)⋮n^2-7\)

\(\Rightarrow n^2+9⋮n^2-7\)

\(\Rightarrow n^2-7-2⋮n^2-7\)

Mà n² - 7 chia hết cho n² - 7

=> \(n^2-7\inƯ(2)\)

\(\Rightarrow n^2-7\in\left\{\pm1;\pm2\right\}\)

Lập bảng :

Vậy \(n\in\left\{3;-3\right\}\)

n.n+2 \(⋮\)n+1

=>\(n^2\)+2\(⋮\)n+1

=>\(n^2\)+2-(n+1)\(⋮\)n+1

=>\(n^2\)+2-n(n+1)\(⋮\)n+1

=>\(n^2\)+2-\(n^2\)-n\(⋮\)n+1

=>2-n\(⋮\)n+1

=>2-n+n+1\(⋮\)n+1

=>3\(⋮\)n+1

=>n+1\(\in\)Ư(3)={\(\mp\)1;\(\mp\)3}

=>n\(\in\){0;-2;2;-4}

Vậy n\(\in\){0;2;-2;-4} thì n.n+2 \(⋮\)n+1

vì n.n+2chia hết cho n+1
ta có:
n.n+2=n^2  +2 =n.(n+1)-n +2=n.(n+1)-(n+1)+1 chia hết cho n+1
mà n.(n+1)-(n+1)chia hết cho n+1
=> 1chia hết cho n+1
=> n+0

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