\(a,\Rightarrow n-1+7⋮n-1\)

Mà \(n-1⋮n-1\Rightarrow7⋮n-1\)

\(\Rightarrow n-1\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow n\in\left\{2;8\right\}\)

\(b,\Rightarrow3\left(n+1\right)+2⋮n+1\)

Mà \(3\left(n+1\right)⋮n+1\Rightarrow2⋮n+1\)

\(\Rightarrow n+1\inƯ\left(2\right)=\left\{1;2\right\}\\ \Rightarrow n=1\left(n\ne0\right)\)

a, 

Ta có: 4n-5 chia hết cho 2n-1

=>4n-2-3 chia hết cho 2n-1

=>2.(2n-1)-3 chia hết cho 2n-1

=>3 chia hết cho 2n-1

=>2n-1=Ư(3)=(-1,-3,1,3)

=>2n=(0,-2,2,4)

=>n=(0,-1,1,2)

Vậy n=0,-1,1,2

b, 

c) 13n⋮n-1

13n-13+13⋮n-1

13n-13⋮n-1 ⇒13⋮n-1

n-1∈Ư(13)

Ư(13)={1;-1;13;-13}

⇒n∈{2;0;14;-12}

b) Bạn tham khảo nha: https://olm.vn/hoi-dap/detail/99050878351.html

a: Ta có: \(n+4⋮n+1\)

\(\Leftrightarrow3⋮n+1\)

\(\Leftrightarrow n+1\in\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow n\in\left\{0;-2;2;-4\right\}\)

mà n là số tự nhiên 

nên \(n\in\left\{0;2\right\}\)

b: Ta có: \(n^2+4⋮n+2\)

\(\Leftrightarrow8⋮n+2\)

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

\(\Leftrightarrow n\in\left\{0;2;6\right\}\)

a: \(\Leftrightarrow7n-7+7⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{2;0;8;-6\right\}\)

b: \(\Leftrightarrow n+1+4⋮n+1\)

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

hay \(n\in\left\{0;-2;1;-3;3;-5\right\}\)

c: \(\Leftrightarrow n^2-9+9⋮n+3\)

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

hay \(n\in\left\{-2;-4;0;-6;6;-12\right\}\)

\(a,\Rightarrow n\inƯ\left(5\right)=\left\{1;5\right\}\\ b,\Rightarrow n\inƯ\left(4\right)=\left\{1;2;4\right\}\\ c,\Rightarrow n\inƯ\left(27\right)=\left\{1;3\right\}\left(n< 7\right)\)

a,( 1;5 )

b, ( 1; 2; 4)

c (1;3 )

n + 5 ⋮ n

=> 5 ⋮ n

=> n thuoc U(5) = {-1; 1; -5; 5}

7n + 8 ⋮ n 

=> 8 ⋮ n 

=> n thuoc U(8) = {-1; 1; -2; 2; -4; 4; -8; 8}

16 - 3n ⋮ n + 4

=> 28 - 3n - 12 ⋮ n + 4

=> 28 - 3(n + 4) ⋮ n + 4

=> 28 ⋮ n + 4

=> n + 4 thuoc U(28) = {-1; 1; -2; 2; -4; 4; -7; 7; -14; 14; -28; 28}

=> n thuoc {-5; -3; -6; -2; -8; 0; -11; 3; -18; 10; -32; 24}

n + 13 ⋮ n - 5

=> n - 5 + 18 ⋮ n - 5

=> 18 ⋮ n - 5

=> n - 5 thuoc U(18) = {-1; 1; -2; 2; -3; 3; -6; 6; -9; 9; -18; 18}

\(n+5⋮n\)

\(\Rightarrow n\inƯ\left(5\right)=\left\{1;5\right\}\)( do \(n\inℕ\))

\(7n+8⋮n\)

có \(7n⋮n\forall n\)\(\Rightarrow\)để \(7n+8⋮n\)cần \(n\inƯ\left(8\right)\)mà \(n\inℕ\Rightarrow n=\left\{1;2;4;8\right\}\)

+) \(3\left(n+1\right)+11⋮n+3\)

\(11⋮n+3\)

\(n+3\inƯ\left(11\right)=\left\{1;11\right\}\)

\(n=8\)

+) \(3n+16⋮n+4\)

\(3\left(n+4\right)+4⋮n+4\)

\(4⋮n+4\)

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

\(n=0\)

+) \(28-7n⋮n+3\)

\(49-7\left(n+3\right)⋮n+3\)

\(49⋮n+3\)

\(n+3\inƯ\left(49\right)=\left\{1;7;49\right\}\)

\(n\in\left\{4;46\right\}\)

g 7n chia het n-3

<=> 7n -21+21 chia het n-3

<=> 7(n-3) +21 chia het n-3

<=> 21 chia het n-3 (vi 7.(n-3) chia het cho n-3)

=> n-3 thuoc uoc cua 21

U(21) ={1;3;7;21}

=>n-3 thuoc{1;3;7;21}

n thuoc {4;6;10;24}

Ta có: n+3 chia hết cho n-1

mà: n-1 chia hết cho n-1

suy ra:[(n+3)-(n-1)]chia hết cho n-1

              (n+3-n+1)chia hết cho n-1

                        4    chia hết cho n-1

                  suy ra n-1 thuộc Ư(4)

           Ư(4)={1;2;4}

suy ra n-1 thuộc {1;2;4}

Ta có bảng sau:

n-1          1             2           4

n              2             3           5

    Vậy n=2 hoặc n=3 hoặc n=5 

Ta có: 2n+1 chia hết cho 2n+1

   nên  2.(2n+1) chia hết cho 2n+1

 suy ra 4n+1 chia hết cho 2n+1

Ta có hiệu sau:

[(4n+3)-(4n+1)] chia hết cho 2n+1

     (4n+3-4n-1) chia hết cho 2n+1

               2     chia hết cho 2n+1

       suy ra  2n+1 thuộc Ư(2)

   Ư(2)={1;2}

suy ra 2n+1∈{1;2}

Ta có bảng sau:

2n+1         1         2

  2n            0        1

   n             0        1/2

    Vậy n=0

a) để n+3⋮n-1

thì n-1+4⋮n-1

⇒4⋮n-1

⇒n-1∈Ư(4)={1;2;4}

\(\Rightarrow\left[{}\begin{matrix}n-1=1\\n-1=2\\n-1=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=2\\n=3\\n=5\end{matrix}\right.\)

vậy n∈{2;3;5}

b)để 4n+3⋮2n+1

thì  2.2n+1+2⋮2n+1

⇒2⋮2n+1

⇒2n+1∈Ư(2)={1;2}

\(\Rightarrow\left[{}\begin{matrix}2n+1=1\\2n+1=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2n=0\\2n=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=0\\n=\dfrac{1}{2}\end{matrix}\right.\)

vì n là số tự nhiên

⇒n=0

vậy n=0

(tick cho mk nha)