Sao câu này giống https://hoc24.vn/cau-hoi/7a-tim-x-z-sao-choa-x-6-chia-het-cho-xb-x-9-chia-het-cho-x-1c-2x-1-chia-het-cho-x-1.3203518129748 thế?

a. x + 6 \(⋮\) x

\(\Rightarrow\left\{{}\begin{matrix}x⋮x\\6⋮x\end{matrix}\right.\)

6 \(⋮\) x

\(\Rightarrow\) x \(\in\) Ư (6) = {1; 2; 3; 6}

\(\Rightarrow\) x \(\in\) {1; 2; 3; 6}

b. x + 9 \(⋮\) x + 1

x + 1 + 8 \(⋮\) x + 1

\(\Rightarrow\left\{{}\begin{matrix}x+1⋮x+1\\8⋮x+1\end{matrix}\right.\)

8 \(⋮\) x + 1

\(\Rightarrow\) x + 1 \(\in\) Ư (8) = {1; 2; 4; 8}

\(\Rightarrow\) x \(\in\) {0; 1; 3; 7}

c. 2x + 1 \(⋮\) x - 1

2x - 2 + 3 \(⋮\) x - 1 

2(x - 1) + 3 \(⋮\) x - 1

\(\Rightarrow\left\{{}\begin{matrix}2\left(x-1\right)⋮x-1\\3⋮x-1\end{matrix}\right.\)

3 \(⋮\) x - 1

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

\(\Rightarrow\) x \(\in\) {2; 4}

a) x Î Ư(6) = {-6; -3; -2; -l; l; 2; 3; 6}.

b) x + l Î Ư (8) = {- 8; -4; -2; -1; 1; 2; 4; 8}. Từ đó tìm được

x Î{-9; -5; -3; -2; 0; 1; 3; 7}.

c)  x - 2 Î Ư(10) = {-10; -5; - 2; -1; 1; 2; 5; 10). Từ đó tìm được

x Î {-8; -3; 0; l; 3; 5; 7; 12}.

.

a) 8 chia hết cho x + 1

--> x + 1 là ước của 8.

TH1: x + 1 = 8 

TH2: x + 1 = 4

TH3: x + 1 = 2

TH4: x + 1 = 1

Giải ra được x = 7; x = 3; x = 1; x = 0

a) \(Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Suy ra \(x\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

b) \(Ư\left(13\right)=\left\{\pm1;\pm13\right\}\)

Suy ra \(x\in\left\{0;12;-2;-14\right\}\)

c) Số nào chia hết cho x - 3 vậy????

d) \(\left(x+8\right)⋮\left(x+2\right)\Leftrightarrow\left(x+2+6\right)⋮\left(x+2\right)\)

Mà x + 2 chia hết cho x + 2 nên 6 chia hết cho x + 2

\(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Suy ra \(x\in\left\{-1;0;1;4;-3;-4;-5;-8\right\}\)

6   \(n^5+5n=n^5-n+6n=n\left(n^4-1\right)+6n=n\left(n^2-1\right)\left(n^2+1\right)+6n\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+6n\)

vì n,n-1 là 2 số nguyên lien tiếp  \(\Rightarrow n\left(n-1\right)⋮2\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\)

  n,n-1,n+1 là 3 sô nguyên liên tiếp \(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮3\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮3\)

\(\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\cdot3=6\)

\(6⋮6\Rightarrow6n⋮6\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)-6n⋮6\Rightarrow n^5+5n⋮6\)(đpcm)

7   \(n\left(2n+7\right)\left(7n+1\right)=n\left(2n+7\right)\left(7n+7-6\right)=7n\left(n+1\right)\left(2n+7\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4+3\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

\(=14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

n,n+1,n+2 là 3 sô nguyên liên tiếp dựa vào bài 6 \(\Rightarrow n\left(n+1\right)\left(n+2\right)⋮6\Rightarrow14n\left(n+1\right)\left(n+2\right)⋮6\)

\(21⋮3;n\left(n+1\right)⋮2\Rightarrow21n\left(n+1\right)⋮3\cdot2=6\)

\(6⋮6\Rightarrow6n\left(2n+7\right)⋮6\)

\(\Rightarrow14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)⋮6\)

\(\Rightarrow n\left(2n+7\right)\left(7n+1\right)⋮6\)(đpcm)

......................?

mik ko biết

mong bn thông cảm 

nha ................

a) 6 ⋮ x

⇒ x ∈ Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}

b) 8 ⋮ (x + 1)

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

⇒ x ∈ {-9; -5; -3; -2; 0; 1; 3; 7}

Vì 6⋮ x

=> x ϵ { 1,2,3,6,-1,-2,-3,-6}

Vì 8 ⋮ x +1

=> x+1 ϵ { 1,2,4,8,-1,-2,-4,-8}

=> x ϵ {0,1,3,7,-2,-3,-5,-9}

=> x = -2