a, x + 2 chia het cho x-1

    x-1 chia het cho x-1 

    => (x+2) - (x-1) chia het cho x-1

    Hay 3 chia het cho x-1

    x-1 thuoc U(3)

    x-1 thuoc {1;3}

    Ta co bang

Vay x thuoc {2;4}

Câu b là dấu nhân hay là x

a)  x + 5 - 8 chia het cho x + 5 

vi x + 5 chia het cho x + 5 

nen  8 chia het cho x + 5 

x + 5 thuoc U(8)={-8;-4;-2;-1;1;2;4;8}

x thuoc{ -13;-9;-7;-6;-4;-3;-1;3}

b )    vi - 10 = 1.(-10)=(-1).10= -2.5=2.(-5)

vi 2x +1 la so chia 2 du 1 

nen 2x + 1 thuoc { -1;1;-5;5}

   x thuoc { -1;0;-3;2}

y - 3 thuoc { -10;10;-2;2}

y thuoc { -7;13;1;5}

tich minh nha ban , thanks

d, ( x+1) nhé. Mình viết nhầm

Trả lời nhanh hộ mình

a)\(\frac{x+11}{x-6}=\frac{x-6+17}{x-6}=\frac{x-6}{x-6}+\frac{17}{x-6}\)

=>x-6\(\in\) Ư(17)

Bài 1:

a/ \(\frac{x+11}{x-6}\)\(\varepsilon\)Z

= \(\frac{x-6+17}{x-6}\)

= \(\frac{x-6}{x-6}\)+ \(\frac{17}{x-6}\)

= \(1+\frac{17}{x-6}\)

\(\Rightarrow\frac{7}{x-6}\)\(\varepsilon\)Z \(\Leftrightarrow x-6\)\(\varepsilon\)\(Ư\left(7\right)\)

\(\Rightarrow x\)\(\varepsilon\)\(\left\{-1;5;7;13\right\}\)

b/ \(\frac{3x+5}{x-2}\)\(\varepsilon\)Z

= \(\frac{3x-6+11}{x-2}\)

= \(\frac{3\left(x-2\right)}{x-2}+\frac{11}{x-2}\)

=  \(3+\frac{11}{x-2}\)

\(\Rightarrow\frac{11}{x-2}\)\(\varepsilon\)Z \(\Leftrightarrow x-2\)\(\varepsilon\)\(Ư\left(11\right)\)

\(\Rightarrow x\)\(\varepsilon\)\(\left\{-5;1;3;9\right\}\)

c/  \(\frac{x^2+11}{x-5}\)\(\varepsilon\)Z

=  \(\frac{x^2-25+36}{x-5}\)

= \(\frac{\left(x+5\right)\left(x-5\right)}{x-5}+\frac{36}{x-5}\)

=  \(x+5+\frac{36}{x-5}\)

Ta có: \(x\)\(\varepsilon\)Z

\(\Rightarrow x+5\)\(\varepsilon\)Z

\(\Rightarrow\frac{36}{x-5}\)\(\varepsilon\)Z \(\Leftrightarrow x-5\)\(\varepsilon\)\(Ư\left(36\right)\)

\(\Rightarrow x\)\(\varepsilon\)\(\left\{-31;-7;-4;-1;1;2;3;4;6;7;8;9;11;14;17;41\right\}\)

Bài 1:a) Ta có: \(1-3x⋮x-2\)

\(\Leftrightarrow-3x+1⋮x-2\)

\(\Leftrightarrow-3x+6-5⋮x-2\)

mà \(-3x+6⋮x-2\)

nên \(-5⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(-5\right)\)

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

hay \(x\in\left\{3;1;7;-3\right\}\)

Vậy: \(x\in\left\{3;1;7;-3\right\}\)

b) Ta có: \(3x+2⋮2x+1\)

\(\Leftrightarrow2\left(3x+2\right)⋮2x+1\)

\(\Leftrightarrow6x+4⋮2x+1\)

\(\Leftrightarrow6x+3+1⋮2x+1\)

mà \(6x+3⋮2x+1\)

nên \(1⋮2x+1\)

\(\Leftrightarrow2x+1\inƯ\left(1\right)\)

\(\Leftrightarrow2x+1\in\left\{1;-1\right\}\)

\(\Leftrightarrow2x\in\left\{0;-2\right\}\)

hay \(x\in\left\{0;-1\right\}\)

Vậy: \(x\in\left\{0;-1\right\}\)

Bài 1 :

a, Có : \(1-3x⋮x-2\)

\(\Rightarrow-3x+6-5⋮x-2\)

\(\Rightarrow-3\left(x-2\right)-5⋮x-2\)

- Thấy -3 ( x - 2 ) chia hết cho  x - 2

\(\Rightarrow-5⋮x-2\)

- Để thỏa mãn yc đề bài thì : \(x-2\inƯ_{\left(-5\right)}\)

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

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

Vậy ...

b, Có : \(3x+2⋮2x+1\)

\(\Leftrightarrow3x+1,5+0,5⋮2x+1\)

\(\Leftrightarrow1,5\left(2x+1\right)+0,5⋮2x+1\)

- Thấy 1,5 ( 2x +1 ) chia hết cho  2x+1

\(\Rightarrow1⋮2x+1\)

- Để thỏa mãn yc đề bài thì : \(2x+1\inƯ_{\left(1\right)}\)

\(\Leftrightarrow2x+1\in\left\{1;-1\right\}\)

\(\Leftrightarrow x\in\left\{0;-1\right\}\)

Vậy ...

\(10^n\)có 1 chữ số 1 và n chữ số 0 nên tổng các chữ số của \(10^n+8\)bằng 9, do vậy nó chia hết cho 9

\(a,x-5⋮x+2\)

\(\Rightarrow x+2-7⋮x+2\)

\(\Rightarrow x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

x + 2 = 1=> x = -1

x + 2 = -1 => x = -3

.... tương tự nhé ~ 

\(2x+3⋮x-5\)

\(\Rightarrow2x-10+7⋮x-5\)

\(\Rightarrow2\left(x-5\right)+7⋮x-5\)

\(\Rightarrow x-5\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

x - 5 = 1 => x = 6 

.... 

lớp 6 rồi nha chú mày