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