Ta có:\(\left(\dfrac{1}{2}x-\dfrac{2}{3}\right)^2=\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{2}{3}\in\left\{\dfrac{1}{2};\dfrac{-1}{2}\right\}\)
TH1:\(\)\(\dfrac{1}{2}x-\dfrac{2}{3}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{1}{2}+\dfrac{2}{3}=\dfrac{7}{6}\)
\(\Rightarrow x=\dfrac{7}{6}:\dfrac{1}{2}=\dfrac{7}{3}\)
TH2:\(\dfrac{1}{2}x-\dfrac{2}{3}=-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{-1}{2}+\dfrac{2}{3}=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}:\dfrac{1}{2}=\dfrac{1}{3}\)
Vậy \(x\in\left\{\dfrac{1}{3};\dfrac{7}{3}\right\}\)
