\(\left(x+3\right)\left(2x-1\right)-\left(x-4\right)\left(2x+1\right)=10\)

\(\Leftrightarrow2x^2-x+6x-3-\left(2x^2+x-8x-4\right)=10\)

\(\Leftrightarrow2x^2-x+6x-3-2x^2-x+8x+4=10\)

\(\Leftrightarrow12x+1=10\)

\(\Leftrightarrow12x=9\)

\(\Leftrightarrow x=\frac{9}{12}\)

\(\Rightarrow x=\frac{3}{4}\)

Vậy   \(x=\frac{3}{4}\)

\(\left|x-\dfrac{1}{2}\right|\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)

\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\\\left|2x-\dfrac{3}{4}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow2x-\dfrac{3}{4}\ge0\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)

\(\Rightarrow\left|x-\dfrac{1}{2}\right|=1\)

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

\(2x-\dfrac{3}{4}\ge0\Rightarrow2x\ge\dfrac{3}{4}\Rightarrow x\ge\dfrac{3}{2}\)

Vậy xảy ra khi:

\(x=\dfrac{3}{2}\)

mk giải cho bạn ở trên r đó

12x/6 + 4x/6 - 3x/6 = 3/8 - 1/4 - 1/6

13x/6 = (9 - 6 - 4)/24 = -1/24

--> x = -1/52

câu trên mk làm rồi

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

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

\(\Rightarrow2x^2-x-2x+1=2x^2-6x+3x-9\)

\(\Rightarrow-x-2x+6x-3x=-1-9\)

\(\Rightarrow0=-10\) (vô lí)

Vậy ko tồn tại giá trị của x.