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\(\left(3n+14\right)⋮\left(n+2\right)\\ \Rightarrow\left[\left(3n+6\right)+8\right]⋮\left(n+2\right)\\ \Rightarrow\left[3\left(n+2\right)+8\right]⋮\left(n+2\right)\)

Vì \(3\left(n+2\right)⋮\left(n+2\right)\Rightarrow8⋮\left(n+2\right)\Rightarrow n+2\in8=\left\{\pm1;\pm2;\pm4;\pm8\right\}\Rightarrow n\in\left\{-10;-6;-4;-3;-1;0;2;6\right\}\)

3n+4=3.(n+2)+2

để 3.(n+2)+2 chia hết cho n+2

=> 2 chia hết cho n+2 

mà n là số tự nhiên

=> n=0

3n + 14 chia hết cho n + 2

⇒ 3n + 6 + 8 chia hết cho n + 2

⇒ 3(n + 2) + 8 chia hết chi n + 2

⇒ 8 chia hết cho n + 2

⇒ n + 2 ∈ Ư(8) = {1; -1; 2; -2; 4; -4; 8; -8}

⇒ n ∈ {-1; -3; 0; -4; 2; -6; 6; -10}

Mà n là số tự nhiên

⇒ n ∈ {0; 2; 6} 

\(\left(3n+14\right)=3\left(n+2\right)+8\)

Để \(\left(3n+14\right)⋮\left(n+2\right)\Rightarrow\left(n+2\right)\inƯ\left(8\right)\)

\(\Rightarrow\left(n+2\right)\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\Rightarrow n\in\left\{-3;-1;-4;0;-6;2;-10;6\right\}\)

yamte aaaa

3n+14 chia hết chi n+1

=>3(n+1)+11 chia hết cho n+1

=>11 chia hết cho n+1

=>n+1 thuộc Ư(11)={1;11}

+)n+1=1=>n=0

+)n+1=11=>n=10

vậy....

n² + 3n + 2 chia hết cho n - 3

=> n² - 3n + 6n - 18 + 20 chia hết cho n - 3

=> (n² - 3n) + (6n - 18) + 20 chia hết cho n - 3

=> n(n - 3) + 6(n - 3) + 20 chia hết cho n - 3

=> (n + 6)(n - 3) + 20 chia hết cho n - 3

Vì (n + 6)(n - 3) chia hết cho n - 3

=> 20 chia hết cho n - 3

=> n - 3 \(\in\)Ư(20) = {-1;1;-2;2-4;4;-5;5;-10;10;-20;20}

=> n \(\in\){2;4;1;5;-1;7;-2;8;-7;13;-17;23}

mày sao lại ra đề khó thế này hả?