\(\dfrac{1}{27}+\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{3}\\ \Leftrightarrow\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{3}-\dfrac{1}{27}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\\ \Leftrightarrow2x-\dfrac{1}{3}=\dfrac{2}{3}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
\(\dfrac{1}{27}\)+\(\left(2x-\dfrac{1}{3}\right)^3\)=\(\dfrac{1}{3}\)
⇔\(\left(2x-\dfrac{1}{3}\right)^3\)=\(\dfrac{1}{3}\)−\(\dfrac{1}{27}\)=\(\dfrac{8}{27}\)=\(\left(\dfrac{2}{3}\right)^3\)
⇔\(2x-\dfrac{1}{3}\)=\(\dfrac{2}{3}\)
⇔2\(x\)=1⇔\(x\)=\(\dfrac{1}{2}\)
Vậy \(x\)=\(\dfrac{1}{2}\)
