Bài 3: 

a chia 36 dư 12 số đó có dạng \(a=36k+12\left(k\in N\right)\)

\(\Rightarrow a=4\left(9k+3\right)\) nên a chia hết cho 4

Mà: \(9k\) ⋮ 3 ⇒ \(9k+3\) không chia hết cho 3

Nên a không chia hết cho 3 

Bài 4:

a) \(x\in B\left(7\right)\) \(\Rightarrow x\in\left\{0;7;14;21;28;35;42;49;...\right\}\)

Mà: \(x\le35\)

\(\Rightarrow x\in\left\{0;7;14;21;28;35\right\}\)

b) \(x\inƯ\left(18\right)\Rightarrow x\in\left\{1;2;3;6;9;18\right\}\)

Mà: \(4< x\le10\)

\(\Rightarrow x\in\left\{6;9\right\}\)

Bài 5:

a) 6 chia hết cho x 

\(\Rightarrow x\inƯ\left(6\right)\)

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

b) \(8\) chia hết cho \(x+1\)

\(\Rightarrow x+1\inƯ\left(8\right)\)

\(\Rightarrow x+1\in\left\{1;2;4;8\right\}\)

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

c) 10 chia hết cho \(x-2\)

\(\Rightarrow x-2\inƯ\left(10\right)\)

\(\Rightarrow x-2\in\left\{1;2;5;10\right\}\)

\(\Rightarrow x\in\left\{3;4;7;12\right\}\)

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: \(x+1\in\left\{1;11\right\}\)

hay \(x\in\left\{0;10\right\}\)

b: \(\Leftrightarrow x+1\in\left\{1;7\right\}\)

hay \(x\in\left\{0;6\right\}\)

a. => \(x\inƯ\left(50\right)=\left\{1;2;5;10;25;50\right\}\)

b. => \(x+1\inƯ\left(32\right)=\left\{1;2;4;8;16;32\right\}\)

=> \(x\in\left\{0;1;3;7;15;31\right\}\)

c. \(\Rightarrow2x+1\inƯ\left(45\right)=\left\{1;3;5;9;15;45\right\}\)

=> \(2x\in\left\{0;2;4;8;14;44\right\}\)

=> \(x\in\left\{0;1;2;4;7;22\right\}\)

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: \(x\in\left\{1;7\right\}\)

b: \(x+1=1\)

hay x=0

\(a,\Rightarrow x\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ b,\Rightarrow2\left(x+1\right)-1⋮x+1\\ \Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Rightarrow x\in\left\{-2;0\right\}\)

a: 

b: 

hay x=0

a) 1 - 2x = 5

2x = -4

x = -2

b) 11 - 5x = 21

5x = -10

x = -2

c) x \(\in\){-13; -1; 1; 13}

a)1080

b)16

c)1;2;3;4;6;12

d)bó tay

e) đề chưa viết xong

k nha

cảm ơnn