\(\Leftrightarrow\sqrt{4x^2-4x+1}=3x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\4x^2-4x+1=\left(3x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\5x^2-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{5}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\dfrac{2}{5}\)
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