1.a)|−7x|=3x+16

Vì |-7x| ≥ 0  nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\)    (*)

Với đk (*), ta có: |-7x|=3x+16

\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔  \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)

⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)

b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)

⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)

⇒ x² - 2x - x + 2 - x² - 2x = 5x - 8  

⇔ -5x - 5x = -8 - 2

⇔ -10x = -10

⇔ x=1

2.7x+5 < 3x−11

⇔ 7x - 3x < -11 - 5

⇔ 4x < -16

⇔ x < -4

bạn tự biểu diễn trên trục số nha !

\(3x-\left(7x+2\right)>5x+4\\ \Leftrightarrow3x-7x-2-5x-4>0\\ \Leftrightarrow-9x-6>0\\ \Leftrightarrow-9x>6\\ \Leftrightarrow x< -\dfrac{2}{3}\)

\(\dfrac{1-2x}{4}-2< \dfrac{1-5x}{8}\)

\(\Leftrightarrow\dfrac{2\left(1-2x\right)-16}{8}< \dfrac{1-5x}{8}\)

\(\Leftrightarrow2\left(1-2x\right)-16< 1-5x\)

\(\Leftrightarrow2-4x-16< 1-5x\)

\(\Leftrightarrow x< 15\)

Vậy \(S=\left\{x|x< 15\right\}\)

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

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

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

=>-5x>=15

=>x<=-3

Trả lời giùm đi mọi người , mình đang cần gấp. Cảm ơn nhiều!!!