\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{3}\\x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)
\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
\(\left(x-\dfrac{1}{3}\right)^2=\perp\left(\dfrac{1}{3}\right)^2\)
\(\text{Vậy }x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x\) \(=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\)
\(\text{hoặc }x-\dfrac{1}{3}=\dfrac{-1}{3}\)
\(x\) \(=\left(\dfrac{-1}{3}\right)+\dfrac{1}{3}=0\)
\(\Rightarrow x\in\left\{\dfrac{2}{3};0\right\}\)
