a: \(\Leftrightarrow n-3\in\left\{-1;1;11\right\}\)

hay \(n\in\left\{2;4;14\right\}\)

Lời giải:

a.

$3n+2\vdots n-3$

$3(n-3)+11\vdots n-3$

$\Rightarrow 11\vdots n-3$

$\Rightarrow n-3\in\left\{1; -1; 11; -11\right\}$

$\Rightarrow n\in\left\{4; 2; 14; -8\right\}$

Vì $n$ tự nhiên nên $n\in\left\{4;2;14\right\}$

b.

$n^2+7n+9\vdots n+7$

$n(n+7)+9\vdots n+7$

$\Rightarrow 9\vdots n+7$

$\Rightarrow n+7\in\left\{1; -1; 3; -3; 9; -9\right\}$

$\Rightarrow n\in\left\{-6; -8; -4; -10; 2; -16\right\}$

Vì $n$ tự nhiên nên $n=2$

a: \(\Leftrightarrow n-3\in\left\{-1;1;11\right\}\)

hay \(n\in\left\{2;4;14\right\}\)

a: \(n^3-2⋮n-2\)

=>\(n^3-8+6⋮n-2\)

=>\(6⋮n-2\)

=>\(n-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(n\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

b: \(n^3-3n^2-3n-1⋮n^2+n+1\)

=>\(n^3+n^2+n-4n^2-4n-4+3⋮n^2+n+1\)

=>\(3⋮n^2+n+1\)

=>\(n^2+n+1\in\left\{1;-1;3;-3\right\}\)

mà \(n^2+n+1=\left(n+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\forall n\)

nên \(n^2+n+1\in\left\{1;3\right\}\)

=>\(\left[{}\begin{matrix}n^2+n+1=1\\n^2+n+1=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n^2+n=0\\n^2+n-2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}n\left(n+1\right)=0\\\left(n+2\right)\left(n-1\right)=0\end{matrix}\right.\Leftrightarrow n\in\left\{0;-1;-2;1\right\}\)

a) 2n+1⋮n-3

2n-6+7⋮n-3

2n-6⋮n-3 ⇒7⋮n-3

n-3∈Ư(7)

Ư(7)={1;-1;7;-7}

⇒n∈{4;2;10;-4}

a. 

Đề bài sai, ví dụ \(n=1\) lẻ nhưng  \(1^2+4.1+8=13\) ko chia hết cho 8

b.

n lẻ \(\Rightarrow n=2k+1\)

\(n^3+3n^2-n-3=n^2\left(n+3\right)-\left(n+3\right)=\left(n^2-1\right)\left(n+3\right)=\left(n-1\right)\left(n+1\right)\left(n+3\right)\)

\(=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1+3\right)\)

\(=8k\left(k+1\right)\left(k+2\right)\)

Do \(k\left(k+1\right)\left(k+2\right)\) là tích 3 số tự nhiên liên tiếp nên chia hết cho 6

\(\Rightarrow8k\left(k+1\right)\left(k+2\right)\) chia hết cho 48