\(\Leftrightarrow1:\left[\dfrac{7}{3}-2x\cdot\dfrac{5}{6}\right]=x-\dfrac{1}{2}\)
=>1:(7/3-5/3x)=x-1/2
=>\(\dfrac{7}{3}-\dfrac{5}{3}x=1:\left(x-\dfrac{1}{2}\right)=1:\dfrac{2x-1}{2}=\dfrac{2}{2x-1}\)
\(\Leftrightarrow\dfrac{5x}{3}=\dfrac{7}{3}-\dfrac{2}{2x-1}=\dfrac{14x-7-6}{3\left(2x-1\right)}=\dfrac{14x-23}{3\left(2x-1\right)}\)
\(\Leftrightarrow5x=\dfrac{14x-23}{2x-1}\)
=>\(10x^2-5x=14x-23\)
=>\(10x^2-19x+23=0\)
hay \(x\in\varnothing\)