Tham khảo 

\(5\left(n+2\right)+4⋮\left(n+2\right)\)

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

Mà \(n\in N\)

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

5(n+2)+4⋮(n+2)5(n+2)+4⋮(n+2)

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

Mà 

5n + 14 = 5n + 10 + 4

= 5(n + 2) + 4

Để (5n + 14) ⋮ (n + 2) thì 4 ⋮ (n + 2)

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

⇒ n ∈ {-6; -4; -3; -2; -1; 0; 2}

Mà n ∈ ℕ

⇒ n ∈ {0; 2}

5n+14 chia hết cho n + 2

=> 5(n+2)+4 chia hết cho n + 2

=> 4 chia hết cho n + 2

=> n + 2 thuộc Ư(4)={1;-1;2;-2;4;-4}

=> n thuộc { -1;-3;0;-4;2;-6}

\(\Leftrightarrow10n+14⋮2n+1\)

\(\Leftrightarrow2n+1\in\left\{1;-1;3;-3;9;-9\right\}\)

\(\Leftrightarrow2n\in\left\{0;-2;2;-4;8;-10\right\}\)

hay \(n\in\left\{0;-1;1;-2;4;-5\right\}\)

5 n chia hết cho n

=>3n chia hết cho n

=> n=1 haowcj 3

Ta có: 5n+10 chia hết cho n-2

=>5n-10+10+10 chia hết cho n-2

=>5.(n-2)+20 chia hết cho n-2

=>20 chia hết cho n-2

=>n-2=Ư(20)=(1,2,4,5,10,20)

=>n=(3,4,6,7,12,22)

Vậy n=3,4,6,7,12,22

a) 3n + 7::5n - 2 

Gọi d là ƯC 3n + 7::5n - 2

\(\hept{\begin{cases}3n+7:d\\5n-2:d\end{cases}}->\hept{\begin{cases}5\left(3n+7\right):d\\3\left(5n-2\right):d\end{cases}}\)

=> 5(3n+7)-3(5n-2):d

      15n+7-15n-2-d

           22n-13n-d

                 9:d=>d=6

b) Tương tự 

\(2x+7=2x+2+5=2\left(x+1\right)+5⋮x+1\\ =>x+1\inƯ\left(5\right)\\ Ư\left(5\right)=\left\{1;-1;5;-5\right\}\\ \left\{{}\begin{matrix}x+1=1\\x+1=-1\\x+1=5\\x+1=-5\end{matrix}\right.=>\left\{{}\begin{matrix}x=0\\x=-2\\x=4\\x=-6\end{matrix}\right.\)