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)

a) \(\left(n+6\right)⋮\left(n+1\right)\Rightarrow\left(n+1\right)+5⋮\left(n+1\right)\)

\(\Rightarrow\left(n+1\right)\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Do \(n\in N\)

\(\Rightarrow n\in\left\{0;4\right\}\)

b) \(\left(4n+9\right)⋮\left(2n+1\right)\Rightarrow2\left(2n+1\right)+7⋮\left(2n+1\right)\)

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

Do \(n\in N\)

\(\Rightarrow n\in\left\{0;3\right\}\)

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, 

\(a,\Rightarrow n+3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow n\in\left\{-8;-4;-2;2\right\}\\ b,\Rightarrow n+3+5⋮n+3\\ \Rightarrow5⋮n+3\\ \Rightarrow n+3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow n\in\left\{-8;-4;-2;2\right\}\\ c,\Rightarrow2\left(2n-1\right)-3⋮2n-1\\ \Rightarrow3⋮2n-1\\ \Rightarrow2n-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Rightarrow n\in\left\{-1;0;1;2\right\}\\ d,\Rightarrow8-n+4⋮8-n\\ \Rightarrow4⋮8-n\\ \Rightarrow8-n\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow n\in\left\{12;10;9;7;6;4\right\}\)

a. n + 4 \(⋮\) n

\(\Rightarrow\left\{{}\begin{matrix}n⋮n\\4⋮n\end{matrix}\right.\)

4 \(⋮\) n 

\(\Rightarrow\) n \(\in\) Ư (4) = {1; 2; 4}

\(\Rightarrow\) n \(\in\) {1; 2; 4}

c. n + 8 \(⋮\) n + 3

n + 3 + 5 \(⋮\) n + 3

\(\Rightarrow\left\{{}\begin{matrix}n+3\text{​​}⋮n+3\\5⋮n+3\end{matrix}\right.\)

\(\Rightarrow\) 5 \(⋮\) n + 3

\(\Rightarrow\) n + 3 \(\in\) Ư (5) = {1; 5}

\(\Rightarrow\) n = 2

b. 3n + 11 \(⋮\) n + 2

3n + 6 + 5 \(⋮\) n + 2

3(n + 2) + 5 \(⋮\) n + 2

\(\Rightarrow\left\{{}\begin{matrix}3\left(n+2\right)\text{​​}⋮n+2\\5⋮n+2\end{matrix}\right.\)

\(\Rightarrow\) 5 \(⋮\) n + 2

\(\Rightarrow\) n + 2 \(\in\) Ư (5) = {1; 5}

\(\Rightarrow\) n = 3

a) \(4\left(n-1\right)-3⋮\left(n-1\right)\)

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

Do \(n\in N\Rightarrow n\in\left\{0;2;4\right\}\)

b) \(-5\left(4-n\right)+12⋮\left(4-n\right)\)

\(\Rightarrow\left(4-n\right)\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)

Do \(n\in N\Rightarrow n\in\left\{16;10;8;7;6;5;3;2;1;0\right\}\)

c) \(-2\left(n-2\right)+6⋮\left(n-2\right)\)

\(\Rightarrow\left(n-2\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Do \(n\in N\Rightarrow n\in\left\{0;1;3;4;5;8\right\}\)

d) \(n\left(n+3\right)+6⋮\left(n+3\right)\)

\(\Rightarrow\left(n+3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Do \(n\in N\Rightarrow n\in\left\{0;3\right\}\)

a) 2n - 4 ⋮ n - 3

2n - 6 + 2 ⋮ n - 3

2( n - 3 ) + 2 ⋮ n - 3

Vì 2( n - 3 ) ⋮ n - 3

=> 2 ⋮ n - 3

=> n - 3 thuộc Ư(2) = { 1; -1; 2; -2 }

=> n thuộc { 4; 2; 5; 1 }

Vậy,......

- Các câu còn lại tương tự

\(a,2n-4⋮n-3\Leftrightarrow2n-6+2⋮n-3\)

\(\Leftrightarrow2\left(n-3\right)+2⋮n-3\Leftrightarrow2⋮n-3\left(n-3\inℤ\right)\)

\(\Leftrightarrow n-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)\(\Leftrightarrow n\in\left\{2;4;1;5\right\}\)

Vậy \(n=1;2;4;5\)

\(b,2n+9⋮n+3\Leftrightarrow2n+6+3⋮n+3\)

\(\Leftrightarrow2.\left(n+3\right)+3⋮n+3\Leftrightarrow3⋮n+3\left(n+3\inℤ\right)\)

\(\Leftrightarrow n+3\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Mà \(n\inℕ\Rightarrow n+3=3\Leftrightarrow n=0\)