Ta có: \(3\cdot\left(\dfrac{1}{3}x-\dfrac{1}{4}\right)^2=\dfrac{1}{3}\)
\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{4}\right)^2=\dfrac{1}{3}:3=\dfrac{1}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{4}=-\dfrac{1}{3}\\\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=-\dfrac{1}{12}\\\dfrac{1}{3}x=\dfrac{7}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{7}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{4};\dfrac{7}{4}\right\}\)
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