2n + 6 = 2n - 2 + 8

Để (2n + 6) ⋮ (2n - 2) thì 8 ⋮ (2n - 2)

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

⇒ 2n ∈{-6; -2; 0; 1; 3; 4; 6; 10}

⇒ n ∈ {-3; -1; 0; 1/2; 3/2; 2; 3; 5}

Cảm ơn nha!!!💐🌷🌷🌷

Bài 1:

Ta có: \(2n^2\left(n+1\right)-2n\left(n^2+n-3\right)\)

\(=2n^3+2n^2-2n^3-2n^2+6n\)

\(=6n⋮6\)

1) \(2n^2\left(n+1\right)-2n\left(n^2+n-3\right)=2n^3+2n^2-2n^3-2n^2+6n=6n⋮6\forall n\in Z\)

2) \(n\left(3-2n\right)-\left(n-1\right)\left(1+4n\right)-1=3n-2n^2-4n^2+3n+1-1=-6n^2+6n=6\left(-n^2+n\right)⋮6\forall n\in Z\)

a, Ta có : \(\text{n + 5 = (n - 1)+6}\)

Vì \(\text{(n-1) ⋮ n-1}\)

Nên để \(\text{n+5 ⋮ n-1}\)⋮ `n-1`

Thì \(\text{6 ⋮ n-1}\) 

\(\Rightarrow\) \(\text{n - 1 ∈ Ư(6)}\)

\(\Rightarrow\) \(\text{n - 1 ∈}\) \(\left\{\text{±1;±2;±3;±6}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{0;-1;-2;-5;2;3;4;7}\right\}\) \(\text{( TM )}\)

\(\text{________________________________________________________}\)

b, Ta có : \(\text{2n-4 = (2n+4)- 8 = 2(n+2) - 8}\)

Vì \(\text{2(n+2) ⋮ n+2}\)

Nên để \(\text{2n-4 ⋮ n+2}\)

Thì \(\text{8 ⋮ n+2}\)

\(\Rightarrow\) \(\text{n + 2 ∈ Ư(8)}\)

\(\Rightarrow\) \(\text{n + 2 ∈}\) \(\left\{\text{±1;±2;±4;±8}\right\}\)

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

\(\text{_________________________________________________________________ }\)

c, Ta có :\(\text{ 6n + 4 = (6n + 3) +1 = 3(2n+1) + 1}\)

Vì \(\text{3(2n+1) ⋮ 2n+1}\)

Nên để\(\text{ 6n+4 ⋮ 2n+1}\)

Thì \(\text{1 ⋮ 2n+1}\)

\(\Rightarrow\) \(\text{2n + 1 ∈ Ư(1)}\)

\(\Rightarrow\) \(\text{2n + 1 ∈}\) \(\left\{\text{±1}\right\}\)

\(\Rightarrow\) \(\text{2n ∈}\) \(\left\{\text{-2;0}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-1;0}\right\}\) ( TM )

\(\text{_______________________________________}\)

Ta có : \(\text{3 - 2n = -( 2n - 3 ) = -( 2n + 2 ) + 5 = -2( n+1)+5}\)

Vì \(\text{-2(n+1) ⋮ n+1}\)

Nên để \(\text{3-2n ⋮ n+1}\)

Thì\(\text{ 5 ⋮ n + 1}\)

\(\Rightarrow\) \(\text{n + 1 ∈}\) \(\left\{\text{±1;±5}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\text{-2;-6;0;4}\) ( TM )

ai tả lời giúp mình với mình đang cần gấp

Trong 3 số `2n+1, 2n+2, 2n+3` luôn có một số chia hết cho 3

\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮3\) (1)

Xét \(n⋮2\)

Có: \(2n⋮2,2⋮2\Rightarrow2n+2⋮2\)

\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮2\) (2)

Xét \(n⋮̸2\)

Có: \(2n⋮2\left(dư1\right),1⋮2\left(dư1\right)\Rightarrow2n+1⋮2\)

\(\Rightarrow\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮2\) (3)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrowđpcm\)

\(2n-1⋮n+1\)

\(\Rightarrow2n+2-3⋮n+1\)

\(\Rightarrow3⋮n+1\)

\(\Rightarrow n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow n+1=1;-1;3;-3\)

\(\Rightarrow n=0;-2;2;-4\)

ta có 10-2n\(⋮\)n-1

\(\Rightarrow\)12-(2n-2)\(⋮\)n-1

mà 2n-2\(⋮\)n-1

\(\Rightarrow\)12\(⋮\)n-1\(\Rightarrow\)n-1\(\in\)Ư(12)={\(\pm\)1;\(\pm\)2;\(\pm\)3;\(\pm\)4;\(\pm\)6;\(\pm\)12)