a) \(\dfrac{15-6x}{3}>5\Leftrightarrow15-6x>15\)

\(\Leftrightarrow-6x>0\Leftrightarrow x< 0\) (vì \(-6< 0\))

\(S=\left\{x|x< 0\right\}\)

b) \(\dfrac{8-11x}{4}< 13\Leftrightarrow8-11x< 52\)

\(\Leftrightarrow-11x< -44\Leftrightarrow x>4\) (vì \(-11< 0\))

\(S=\left\{x|x>4\right\}\)

c) \(8x+3\left(x+1\right)>5x-\left(2x-6\right)\)

\(\Leftrightarrow8x+3x+1>5x-2x+6\)

\(\Leftrightarrow8x+3x-5x+2x>6-1\)

\(\Leftrightarrow8x>5\)

\(\Leftrightarrow x>\dfrac{5}{8}\) (vì \(8>0\))

\(S=\left\{x|x>\dfrac{5}{8}\right\}\)

d) \(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\)

\(\Leftrightarrow12x^2-2x-12x^2-9x+8x>-6\)

\(\Leftrightarrow-3x>-6\)

\(\Leftrightarrow x< 2\) (vì \(-3< 0\))

\(S=\left\{x|x< 2\right\}\)

a) \(\dfrac{15-6x}{3}>5\) <=> \(15-6x>15\) <=> \(6x< 0\) <=> \(x< 0\)

b) \(\dfrac{8-11x}{4}< 13\) <=> \(8-11x< 52\) <=> \(11x>-44\)<=> \(x>-4\)

c) \(8x+3\left(x+1\right)>5x-\left(2x-6\right)\)

<=> 8x + 3x + 3 - 5x + 2x - 6 > 0

<=> 8x  > 3

<=> x > 3/8

d) 2x(6x - 1) > (3x - 2)(4x + 3)

<=> 12x² - 2x > 12x² + x - 6

<=> 12x² - 2x - 12x² - x > -6

<=> -3x > -6

<=> x < 2

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(ĐKXĐ:x\ne-1,x\ne3\right)\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}\)

\(\Rightarrow x\left(x+1\right)-x\left(x-3\right)=4x\)

\(\Leftrightarrow x^2+x-x^2+3x=4x\)

\(\Leftrightarrow x^2+x-x^2+3x-4x=0\)

\(\Leftrightarrow0x=0\)

Phương trình có vô số nghiệm , trừ x = -1,x = 3

Vậy ...

\(\dfrac{12x+1}{12}< \dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow12\cdot\dfrac{12x+1}{12}< 12\cdot\dfrac{9x+1}{3}-12\cdot\dfrac{8x+1}{4}\)

\(\Leftrightarrow12x+1< 4\left(9x+1\right)-3\left(8x+1\right)\)

\(\Leftrightarrow12x+1< 36x+4-24x-3\)

\(\Leftrightarrow12x+1< 12x+1\)

\(\Leftrightarrow12x-12x< 1-1\)

\(\Leftrightarrow0x< 0\)

Vậy S = {x | x \(\in R\)}

roois vãi

-Đăng tách câu hỏi bạn nhé.

a) \(8x+3\left(x+1\right)>5x-\left(2x-6\right)\)

⇒ \(8x + 3x + 3 > 5x - 2x + 6\)

⇒ \(11x+3>3x+6\)

⇒ \(11x - 3x > 6 -3\)

⇒ \(8x > 3\)

⇒ \(8x.\dfrac{1}{8}>3.\dfrac{1}{8}\)

⇒ \(x>\dfrac{3}{8}\)

S = \(\left\{x\backslash x>\dfrac{3}{8}\right\}\)

b) \(2x(6x-1) > (3x -2)(4x+3)\)

⇒ \(12x^2 - 2x > 12x^2 +9x -8x -6\)

⇒ \(12x^2 - 2x > 12x^2 + x - 6\)

⇒ \(-2x-x>12x^2 -6-12x^2\)

⇒ \(- 3x > -6 \)

⇒ \(x > 2\)

S = {x / x > 2}

`b,2(x+1)=5x-7`

`=>2x+2=5x-7`

`=>3x=9`

`=>x=3`

`d,(10x+3)/12=1+(6+8x)/9`

`<=>(10x+3)/12=(8x+15)/9`

`<=>30x+9=32x+60`

`<=>2x=-51`

`<=>x=-51/2`

\(\dfrac{2x-3}{2}>\dfrac{8x-11}{6}\)

\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}>\dfrac{8x-11}{6}\)

\(\Leftrightarrow3\left(2x-3\right)>8x-11\)

\(\Leftrightarrow6x-9>8x-11\)

\(\Leftrightarrow-2x>-2\)

\(\Leftrightarrow x< 1\)

Vậy \(S=\left\{x|x< 1\right\}\)

\(2x-3\le8x-11\)

\(\Leftrightarrow-6x\le-8\)

\(\Leftrightarrow x\ge\dfrac{8}{6}\)

Vậy \(S=\left\{x|x\ge\dfrac{8}{6}\right\}\)

a: Ta có: \(3x+5\le4x-9\)

\(\Leftrightarrow-x\le-14\)

\(\Leftrightarrow x\ge14\)

b: Ta có: \(6-2x< 6-x\)

\(\Leftrightarrow-x< 0\)

hay x>0

c: Ta có: \(7\left(x-1\right)+5>-3x\)

\(\Leftrightarrow7x-7+5+3x>0\)

\(\Leftrightarrow10x>2\)

hay \(x>\dfrac{1}{5}\)