\(\left|2x-1\right|-x=4\Leftrightarrow\left|2x-1\right|=4+x\)          (1)

+)TH1: \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\)   thì ph(1) trở thành

    \(2x-1=4+x\Leftrightarrow x=5\) (tm)

+)TH2: \(2x-1< 0\Leftrightarrow x< \frac{1}{2}\) thì pt(1) trở thành

  \(1-2x=4+x\Leftrightarrow-3x=3\Leftrightarrow x=-1\) (tm)

Vậy x={-1;5}

Vẫn thức à

|x-3|+1=x

|x+3| = x-1

Th1: x+3 = x-1 

0x = 4

x thuộc rỗng

Th2: -(x+3) = x-1

-x -3 = x-1

2x = -2

x= -1 

Vậy x=-1

|x-3|=2x+4

Th1: x-3 = 2x +4

x = -7

Th2: -(x-3) = 2x+4

-x +3 = 2x +4

-3x = 1

x= -1/3 

Vậy x= -7; x= -1/3

|3-2x|=x+1

Th1: 3 - 2x = x+1 

3x = 2

x= 2/3

Th2: - (3-2x) = x+1

- 3 +2x = x+1

x= 4

Vậy x= 2/3 và x= 4

|x-1|+3x=1

|x+1| = 1- 3x

Th1: x+1 = 1- 3x 

4x = 0

x=0 

Th2: -(x+1) = 1- 3x

-x -1 = 1- 3x

2x = 2

x=1

Vậy x=0 và x=1

bạn có thể giải thích rõ ra tí được ko bạn 

sao bạn ko trả lời 

\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(\left(x+3\right)\left(2x-4\right)< 0\)

\(\Rightarrow2\left(x+3\right)\left(x-2\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+3>0\\x-2< 0\end{matrix}\right.\\\left[{}\begin{matrix}x+3< 0\\x-2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x>-3\\x< 2\end{matrix}\right.\\\left[{}\begin{matrix}x< -3\\x>2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>-3\\x< 2\end{matrix}\right.\)

\(\Rightarrow-3< x< 2\)

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

\(\left(x+\frac{2}{3}\right)\left(\frac{5}{4}-2x\right)>0\)

th1 : 

\(\hept{\begin{cases}x+\frac{2}{3}>0\\\frac{5}{4}-2x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-\frac{2}{3}\\-2x>-\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x>-\frac{2}{3}\\x>\frac{5}{8}\end{cases}\Rightarrow}x>\frac{5}{8}}\)

th2 : 

\(\hept{\begin{cases}x+\frac{2}{3}< 0\\\frac{5}{4}-2x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -\frac{2}{3}\\-2x< -\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x< -\frac{2}{3}\\x< \frac{5}{8}\end{cases}\Rightarrow}x< -\frac{2}{3}}\)

vậy_