\(2x+10⋮x+1\)

\(\Rightarrow2x+2+8⋮x+1\)

\(\Rightarrow2\left(x+1\right)+8⋮x+1\)

\(\Rightarrow x+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(x\in\left\{0;1;-2;\pm3;-5;7;-9\right\}\)

\(2x+10⋮x-1\)

\(\Rightarrow2x-2+12⋮x-1\)

\(\Rightarrow2\left(x-1\right)+12⋮x-1\)

\(\Rightarrow x-1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

x - 1 = 1 => x = 2

x -1 = -1 => x = 0 

... tg tự 

\(3x+10⋮x+1\)

\(\Rightarrow3x+3+7⋮x+1\)

\(\Rightarrow3\left(x+1\right)+7⋮x+1\)

\(\Rightarrow x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

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

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) x = 1

2) x = 7

3) x = 9

4) x = 3

5) x = 3

k mik nhé

Ta có : 10⋮(x+1)=>(x+1)ϵƯ(10)={1;-1;2-2;5;-5;10;-10}

Ta tìm x theo bảng giá trị sau:

XOG RÙI ĐÓ. CHÚC THI TỐT NHA

Còn câu b)&c) thì làm tương tự nhé.

Chị học lớp chuyên Toán nên cứ hỏi

Câu b)mỏi tay lắm

Đề gì bn

Là sao đề đâu >??

ϡϡϡϡϡϡϡϡϡϡ

b/10 chia hết cho 3x+1

=>3x+1 thuộc Ư(10)={1;2;5;10}

3x+1=1=>3x=0=>x=0

3x+1=2=>3x=1=>x=1/3 

3x+1=5=>3x=4=>x=4/3

3x+1=10=>3x=9=>x=3

vì x thuộc N nên x ={0;3}

c) x+16 chia hết cho x+1

=>(x+1)+15 chia hết cho x+1

=>x+1 thuộc Ư(15)={1;3;5;15}

x+1=1=>x=0

x+1=3=>x=2

x+1=5=>x=4

x+1=15=>x=14

=>x thuộc {0;2;4;14}

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: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

hay \(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)