5.

$4x+3\vdots x-2$

$\Rightarrow 4(x-2)+11\vdots x-2$

$\Rightarrow 11\vdots x-2$

$\Rightarrow x-2\in \left\{1; -1; 11; -11\right\}$

$\Rightarrow x\in \left\{3; 1; 13; -9\right\}$

6.

$3x+9\vdots x+2$
$\Rightarrow 3(x+2)+3\vdots x+2$
$\Rightarrow 3\vdots x+2$

$\Rightarrow x+2\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow x\in \left\{-1; -3; 1; -5\right\}$

7.

$3x+16\vdots x+1$

$\Rightarrow 3(x+1)+13\vdots x+1$

$\Rightarrow 13\vdots x+1$

$\Rightarrow x+1\in \left\{1; -1; 13; -13\right\}$

$\Rightarrow x\in\left\{0; -2; 12; -14\right\}$

8.

$4x+69\vdots x+5$

$\Rightarrow 4(x+5)+49\vdots x+5$

$\Rightarrow 49\vdots x+5$

$\Rightarrow x+5\in\left\{1; -1; 7; -7; 49; -49\right\}$

$\Rightarrow x\in \left\{-4; -6; 2; -12; 44; -54\right\}$

** Bổ sung điều kiện $x$ là số nguyên.

1. $x+9\vdots x+7$

$\Rightarrow (x+7)+2\vdots x+7$

$\Rightarrow 2\vdots x+7$

$\Rightarrow x+7\in \left\{1; -1; 2; -2\right\}$

$\Rightarrow x\in \left\{-6; -8; -5; -9\right\}$

2. Làm tương tự câu 1

$\Rightarrow 9\vdots x+1$

3. Làm tương tự câu 1

$\Rightarrow 17\vdots x+2$
4. Làm tương tự câu 1

$\Rightarrow 18\vdots x+2$

1) \(\left(x+8\right)⋮x\)

\(\Rightarrow x+8-x⋮x\)

\(\Rightarrow8⋮x\)

\(\Rightarrow x\in U\left(8\right)=\left\{-1;1;-2;2;-4;4;-8;8\right\}\left(x\in Z\right)\)

2) \(3x+16⋮x+4\)

\(\Rightarrow3x+16-3\left(x+4\right)⋮x+4\)

\(\Rightarrow3x+16-3x-12⋮x+4\)

\(\Rightarrow4⋮x+4\)

\(\Rightarrow x+4\in U\left(4\right)=\left\{-1;1;-2;2;-4;4\right\}\)

\(\Rightarrow x\in=\left\{-5;-3;-6;-2;-8;0\right\}\left(x\in Z\right)\)

Bài này dể mà ko bít lm haizz

a: \(2x+1⋮x-3\)

\(\Leftrightarrow2x-6+7⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{4;2;10;-4\right\}\)

b: \(\Leftrightarrow x+2-17⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;17;-17\right\}\)

hay \(x\in\left\{-1;-3;15;-19\right\}\)

c: \(\Leftrightarrow3x+3+13⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1;13;-13\right\}\)

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

d: \(\Leftrightarrow3x-6+1⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1\right\}\)

hay \(x\in\left\{3;1\right\}\)

e: \(\Leftrightarrow5x+5-8⋮x+1\)

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

hay \(x\in\left\{0;-2;1;-3;3;-5;7;-9\right\}\)

Đề dài quá làm không nổi ... Làm mẫu 1 - 2 ý thôi nhá

2x + 1 chia hết cho x - 3

=> 2(x - 3) + 7 chia hết cho x - 3

=> 2x - 6 + 7 chia hết cho x -3

=> 7 chia hết cho x - 3

=> x - 3 thuộc Ư(7) = { -7 ; -1 ; 1 ; 7 }

x - 15 chia hết cho x + 2

=> x + 2 - 17 chia hết cho x + 2

=> 17 chia hết cho x + 2

=> x + 2 thuộc Ư(17) = { -17 ; -1 ; 1 ; 7 }

Các ý còn lại làm tương tự

a: \(3x+1\in\left\{1;10;2;5\right\}\)

\(\Leftrightarrow3x\in\left\{0;9;1;4\right\}\)

hay \(x\in\left\{0;3;\dfrac{1}{3};\dfrac{4}{3}\right\}\)

b: \(x+3\in\left\{3;4;6;12\right\}\)

hay \(x\in\left\{0;1;3;9\right\}\)

a) x=3

b) x=2

kết quả của bài này là:

a) x = 3

b) x = 2

nhớ ấy cho mình nhé

a)\(10⋮\left(3x+1\right)\)

\(\Rightarrow3x+1\inƯ\left(10\right)=\left\{1;2;5;10\right\}\)

Ta có bảng sau :

Vậy \(x=3\)

b) \(\left(x+16\right)⋮\left(x+1\right)\)

\(\left(x+1+15\right)⋮\left(x+1\right)\)

\(\Rightarrow\hept{\begin{cases}x+1⋮x+1\\15⋮x+1\end{cases}}\)

\(\Rightarrow x+1\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)

Ta có bảng sau :

Vậy \(x\in\left\{0;2;4;15\right\}\)

( 3x + 16 )\(⋮\)( x + 2 )

\(\Rightarrow\)3x + 16 = 3 ( x + 2 ) +10

\(\Rightarrow\)10\(⋮\)( x + 2 )

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

\(\Rightarrow\)x\(\in\left\{0;3;8\right\}\)

\(\left(3x+16\right)⋮x+2\)

\(\Rightarrow3\left(x+2\right)+10⋮x+2\)

\(\Rightarrow10⋮x+2\Rightarrow x+2\inƯ\left(10\right)\)

Mà \(Ư\left(10\right)=\left\{1;2;5;10\right\}\)

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

\(\Rightarrow x\in\left\{-1;0;3;8\right\}\)

Vì \(x\in N\Rightarrow x\in\left\{0;3;8\right\}\)

( 3x + 16 )\(⋮\)( x + 2 ) \(\Rightarrow\)3x + 16 = 3 ( x + 2 ) +10 \(\Rightarrow\)10\(⋮\)( x + 2 ) \(\Rightarrow\)x + 2 \(\in\)\(\left\{1;2;5;10\right\}\) \(\Rightarrow\)x\(\in\left\{0;3;8\right\}\)\(( 3x + 16 )\(⋮\)( x + 2 ) \(\Rightarrow\)3x + 16 = 3 ( x + 2 ) +10 \(\Rightarrow\)10\(⋮\)( x + 2 ) \(\Rightarrow\)x + 2 \(\in\)\(\left\{1;2;5;10\right\}\) \(\Rightarrow\)x\(\in\left\{0;3;8\right\}\)\)