\(-\frac{1}{4}x+\frac{3}{2}x-\frac{2}{3}x+6=\)\(0\)

\(\Rightarrow\)\(-\frac{1}{4}x+\frac{3}{2}x-\frac{2}{3}x\)\(=-6\)

\(\Rightarrow\)\(x\left(-\frac{1}{4}+\frac{3}{2}-\frac{2}{3}\right)\)\(=-6\)

\(\Rightarrow\)\(x.\frac{7}{12}\)\(=-6\)

\(\Rightarrow\)\(x\)\(=-\frac{72}{7}\)

\(\text{Học tốt!!!}\)

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

Bài làm:

a) \(2\left|x-1\right|-8=0\)

\(\Leftrightarrow2\left|x-1\right|=8\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)

b) \(-\left|2x+3\right|+3=6\)

\(\Leftrightarrow\left|2x+3\right|=-3\)

Mà \(\left|2x+3\right|\ge0>-3\left(\forall x\right)\)

=> Mâu thuẫn

=> Không tồn tại x thỏa mãn

a) Ta có 2|x - 1| - 8 = 0

=> 2|x - 1| = 8

=> |x - 1| = 4

=> \(\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)

b) Ta có : -|2x + 3| + 3 = 6

=> -|2x + 3| = 3

=> |2x + 3| = -3

Vì \(\left|2x+3\right|\ge0\forall x\)

mà -3 < 0

=> x \(\in\varnothing\)

a, \(2\left|x-1\right|-8=0\Leftrightarrow2\left|x-1\right|=8\Leftrightarrow\left|x-1\right|=4\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)

b,\(-\left|2x+3\right|+3=6\Leftrightarrow-\left|2x+3\right|=3\Leftrightarrow\left|2x+3\right|=-3\)

Vì \(\left|2x+3\right|\ge0\forall x\)Mà \(-3< 0\)

\(\Rightarrow x\in\varnothing\)

a) \(\left(x-\frac{1}{2}\right)\left(2x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\2x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)

b) \(\left(x-\frac{1}{2}\right)\left(x+2\right)< 0\)

TH1: \(\hept{\begin{cases}x-\frac{1}{2}< 0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}x< \frac{1}{2}\\x< -2\end{cases}}}\)

TH2: \(\hept{\begin{cases}x-\frac{1}{2}>0\\x+2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>\frac{1}{2}\\x< -2\end{cases}}}\)

cho mình hỏi th1 và th 2 là gi vậy

Lời giải:

$(\frac{1}{2}+2x)(2x-3)=0$

$\Leftrightarrow \frac{1}{2}+2x=0$ hoặc $2x-3=0$

$\Rightarrow x=\frac{-1}{4}$ hoặc $x=\frac{3}{2}$

(\(\dfrac{1}{2}\) + 2\(x\))(2\(x\) - 3) =0

\(\left[{}\begin{matrix}\dfrac{1}{2}+2x=0\\2x-3=0\end{matrix}\right.\)

  \(\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=3\end{matrix}\right.\)

   \(\left[{}\begin{matrix}x=-\dfrac{1}{2}:2\\x=3:2\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(\left(\dfrac{1}{2}+2x\right).\left(2x-3\right)=0\)

- Ta có 2 trường hợp:

 \(\dfrac{1}{2}+2x=0\) ⇒ \(x=-\dfrac{1}{4}\)

\(2x-3=0\) ⇒ \(x=1,5\)