\(6x-1\in2x-1\)

\(2x-1⋮6x-1\)

\(3\left(2x-1\right)⋮6x-1\)

\(6x-3⋮6x-1\)

\(6x-1-2⋮6x-1\)

\(-2⋮6x-1\)

\(\Rightarrow6x-1\inƯ\left(-2\right)=\left\{\pm1;\pm2\right\}\)

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x²-x-4x+2-2x²-6x=-11x+2

Bài 1:

a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)

\(A=2x^2-x-4x+2-2x^2-6x\)

\(A=-11x+2\)

b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)

\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)

\(B=-12x\)

c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

Bài 2: 

a: Ta có: \(2x\left(3x-5\right)\left(x+11\right)-3x\left(2x+3\right)\left(x+7\right)\)

\(=2x\left(3x^2+33x-5x-55\right)-3x\left(2x^2+14x+3x+21\right)\)

\(=6x^3+56x^2-110x-6x^2-51x^2-63x\)

\(=-117x\)

b: Ta có: \(\left(x^2+5x-6\right)\left(x-1\right)-\left(x+2\right)\left(x^2-x+1\right)-x\left(3x-10\right)\)

\(=x^3+4x^2-11x+6-\left(x^3-x^2+x+2x^2-2x+2\right)-3x^2+10x\)

\(=x^3+x^2-x+6-x^3-x^2+x-2\)

=4

c: Ta có: \(\left(x^2+x+1\right)\left(x-1\right)-x^2\left(x+1\right)+x^2-5\)

\(=x^3-1-x^3-x^2+x^2-5\)

=-6

ghi rõ các bược hộ em ạ 

a/ (2x + 1)(4x – 3) – 6x(x + 5) – 2x(x – 7) + 18x

=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x

=-3

vậy...

a) Ta có: \(\left(2x+1\right)\left(4x-3\right)-6x\left(x+5\right)-2x\left(x-7\right)+18x\)

\(=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x\)

\(=-3\)

b) Ta có: \(9x\left(2x-5\right)-\left(6x+2\right)\left(3x-2\right)+39x\)

\(=18x^2-45x-18x^2+12x-6x+4+39x\)

\(=4\)

c) Ta có: \(4x\left(2x-3\right)+x\left(x+2\right)-9x\left(x-1\right)+x-5\)

\(=8x^2-12x+x^2+2x-9x^2+9x+x-5\)

\(=-5\)

sai đề r

Bài làm

Ta có: 6x - 1 thuộc 2x - 1

=> 2x  - 1 chia hết chp 6x - 1

=> 6x - 3 chia hết cho 6x - 1

=> -2 chia hết cho 6x - 1

=> 6x - 1 thuộc Ư_(-2) = { 1; -1; 2; -2 }

Ta có bảng sau:

Vậy x = { \(\frac{1}{3}\); 0; \(\frac{1}{2}\); \(-\frac{1}{6}\)} 

Ta có:

\(\left(6x+2\right)⋮\left(2x-1\right)\Leftrightarrow\left[3\left(2x-1\right)+5\right]⋮\left(2x-1\right)\)

\(\Leftrightarrow5⋮\left(2x-1\right)\Leftrightarrow2x-1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow x\in\left\{-2;0;1;3\right\}\)

trả lời nhanh giúp mình

\(\Rightarrow3\left(2x-1\right)+5⋮\left(2x-1\right)\\ \Rightarrow2x-1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow x\in\left\{-2;0;1;3\right\}\)

\(A=\dfrac{6x+5}{2x-1}=\dfrac{3\left(2x-1\right)+8}{2x-1}=3+\dfrac{8}{2x-1}\)

\(\Rightarrow2x-1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(6x-9⋮2x+1\)

\(\Rightarrow3\left(2x+1\right)-12⋮2x+1\)

\(\Rightarrow12⋮2x+1\)

Tới đây bí=)alibaba!

Forever Mis You tìm tất cả các ước của 12 (cả âm và dương) rồi tính từng trường hợp một nha 

\(6x-9⋮2x+1\)

\(\Rightarrow3\left(2x+1\right)-12⋮2x+1\)

\(\Rightarrow12⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Vì 2x+1 chia cho 2 dư 1 

\(\Rightarrow2x+1\in\left\{-1;1;3;-3\right\}\)

\(\Rightarrow x\in\left\{-1;0;1;-2\right\}\)

Vậy.................

 a,2x(3x-1)-6x(x+1)-(3-8x)

=6x^2-2x-6x^2-6x-3+8x

=-3

Vậy............

Ta có 2x(3x - 1) - 6x(x  +1) - (3 - 8x) 

= 6x² - 2x - 6x² - 6x - 3 + 8x 

= (6x² - 6x²) + (8x - 2x - 6x) - 3 = - 3

=> Biểu thức trên không phụ thuộc vào biến