De 6n+7 chia het cho 2n-1

thi 6n+7 chia het cho 2n-1 va 2n-1 chia het cho 2n-1

=> 6n+7 chia het cho 2n-1 va 3.(2n-1) chia het cho 2n-1

=> 6n+7 chia het cho 2n-1 va 6n-3 chia het cho 2n-1

=> (6n+7)-(6n-3) chia het cho 2n-1

=> 6n+7-6n+3 chia het cho 2n-1

=> 10 chia het cho 2n-1

=> 2n-1 thuoc U(10)={1, -1, 2, -2, 5, -5, 10, -10}

phan con lai ban tu lam tiep nhe

Vì 6n + 7 ⋮ 2n - 1 ⇒ 2n + 2n + 2n - 1 - 1 - 1 + 10 ⋮ 2n  1

 ⇒ ( 2n - 1 ) + ( 2n - 1 ) + ( 2n - 1 ) + 10 ⋮ 2n - 1

Vì 2n - 1 ⋮ 2n - 1 . Để ( 2n - 1 ) + ( 2n - 1 ) + ( 2n - 1 ) + 10 ⋮ 2n - 1 ⇒ 10 ⋮ 2n - 1

⇒ 2n - 1 ∈ Ư ( 10 )

⇒ Ư ( 10 ) = { + 1 ; + 2 ; + 5 ; + 10 }

⇒ 2n - 1 = + 1 ; + 2 ; + 5 ; + 10

⇒ 2n = 2 ; 0 ; 3 ; - 1 ; 6 ; - 4 ; 11 ; - 9

 ⇒ n = 1 ; 0 ; 3 ; - 2 

 Vậy n = { - 2 ; 0 ; 1 ; 3 }

n+ 9 \(⋮n-2\)

mà n - 2 \(⋮n-2\)

= n -2 +11 \(⋮n-2\)

=> 11 \(⋮n-2\)

n -2 \(\inư\left(11\right)\in1,11\)

Ta có bảng: 

Vậy x = 3; 13

Thanks bạn nha !!!!!!!!!!! Kb vs mk nha!!!!!!!!!!!!

=>(n²+3n)+(3n+9)+2 chia hết cho n+3

=>n(n+3)+3(n+3)+2 chia hết cho n+3

=>(n+3)(n+3)+2 chia hết cho n+3

Mà (n+3)(n+3) chia hết cho n+3

=>2 chia hết cho n+3

=> n+3 thuộc Ư(2)={1;2;-1;-2}

=>n thuộc {-2;-1;-4;-5}

Để A nguyên

=>n²-3n+1 chia hết cho n+1

=>(n²-1)-(3n+3)+1+1-3 chia hết cho n+1

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

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

=>1 chia hết cho n+1

=>n+1 thuộc Ư(1)={1;-1}

=>n thuộc {0;-2}

bài nào z bn

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 )

a) \(3n+19⋮n+1\)

\(\Rightarrow\)\(3\left(n+1\right)+16⋮n+1\)

mà \(3\left(n+1\right)⋮n+1\)\(\Rightarrow\)\(16⋮n+1\)

\(\Rightarrow\)\(n+1\in\left\{1,-1,2,-2,4,-4,8,-8,16,-16\right\}\)

\(\Rightarrow n\in\left\{0,-2,1,-3,3,-5,7,-9,15,-17\right\}\)

b) \(2n+7⋮n+2\)

\(\Rightarrow2\left(n+2\right)+3⋮n+2\)

mà \(2\left(n+2\right)⋮n+2\Rightarrow3⋮n+2\)

\(\Rightarrow n+2\in\left\{1,3,-1,-3\right\}\)

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

c)\(6n+39⋮2n+1\Rightarrow3\left(2n+1\right)+36⋮2n+1\)

mà\(3\left(2n+1\right)⋮2n+1\)\(\Rightarrow36⋮2n+1\)

\(\Rightarrow2n+1\in\left\{1,-1,2,-2,3,-3,4,-4,6,-6,9,-9,12,-12,18,-18,36,-36\right\}\)

\(\Rightarrow2n\in\left\{0,-2,1,-3,2,-4,3,-5,5,-7,8,-10,11,-13,17,-19,35,-37\right\}\)

\(\Rightarrow\)\(n\in\left\{0,-1,1,-2,4,-5\right\}\)