\(\text{Ta có}:\)

\(27-5n\)\(⋮\)\(n\)

\(\text{Mà 5n}\)\(⋮\)\(5\)\(\text{nên 27 }⋮\)\(5\)

\(\Rightarrow n\inƯ\left(27\right)=\left\{1;3;9;27\right\}\)

\(\text{Vậy}\)\(n\in\left\{1;3;9;27\right\}\)

\(\text{Ta có}:\)

\(n+6\)\(⋮\)\(n+2\)

\(\Rightarrow n+2+4\)\(⋮\)\(n+2\)

\(\text{Mà n + 2}\)\(⋮\)\(n+2\)\(\text{nên 4}\)\(⋮\)\(n+2\)

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

\(\Rightarrow n\in\left\{-1,-3,0,-5,2,-6\right\}\)

\(\text{Mà n }\in N\)\(\text{nên}\)\(:\)

\(n\in\left\{0,2\right\}\)

ta có 7-5n  n+2

     5n+10  n+2

=>7-5n+5n+10  n+2

=>17  n+2

=>n+2{1;17;-1;-17}

=>n{-1;15;-3;-19}

nha, cảm ơn!

11,

a, 4x-3\(\vdots\) x-2 ¹

    x-2\(\vdots\) x-2\(\Rightarrow\) 4(x-2)\(\vdots\) x-2\(\Rightarrow\) 4x-8\(\vdots\) x-2 ²

Từ ¹ và ² ta có:

(4x-3)-(4x-8)\(\vdots\) x-2

\(\Rightarrow\) 4x-3-4x+8\(\vdots\) x-2

\(\Rightarrow\)       5       \(\vdots\) x-2

\(\Rightarrow\) x-2\(\in\) Ư(5)

\(\Rightarrow\) x-2\(\in\){-5;-1;1;5}

\(\Rightarrow\) x\(\in\) {-3;1;3;7}

Vậy......

Phần b và c làm tương tự như phần a pn nhé!

n+1\(⋮\)7 

\(\Rightarrow\)5n+1+14\(⋮7\)

\(\Rightarrow5n+15⋮7\)

\(\Rightarrow5(n+3)⋮7\)

\(\Rightarrow n+3⋮7\left(vi(5:7)=1\right)\)

\(\Rightarrow n+3\in B_{\left(7\right)}\)

\(\Rightarrow n+3=7k\left(k\inℕ^∗\right)\)

\(\Rightarrow n=7k-3\)

vậy n có dạng 7k-3

a, Ta có : 2n + 19 chia hết cho 7

\(\Rightarrow\) \(2n+19\inƯ\left(7\right)\)

Mà  \(Ư\left(7\right)=\left\{1;-1;7;-7\right\}\)

\(\Rightarrow\) \(2n+19\in\left\{1;-1;7;-7\right\}\)

\(\Rightarrow\) \(2n\in\left\{20;18;26;12\right\}\)

\(\Rightarrow\) \(n\in\left\{10;9;13;6\right\}\)

a.n=3

b.n=55

a,Ta có : 2n+7= 2(n-3)+7

          Mà 2n+7 ⋮  n-3 => 2(n-3) ⋮ n-2 => 7 ⋮ n-3 => n-3 ∈ Ư(7)

          => n-3∈{1,7,-1,-7} => n ∈ { 4,10,2,-4}

b, câu b sai đề, bạn xem lại nhé