\(\sqrt{4x^2-4x+1}=\sqrt{3}-1\left(1\right)\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{3}-1\)
\(\Leftrightarrow\left|2x-1\right|=\sqrt{3}-1\)
TH1: \(x\ge\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow2x-1=\sqrt{3}-1\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
TH2: \(x< \dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow1-2x=\sqrt{3}-1\)
\(\Leftrightarrow x=\dfrac{2-\sqrt{3}}{2}\)
Ta có: \(\sqrt{4x^2-4x+1}=\sqrt{3}-1\)
\(\Leftrightarrow\left|2x-1\right|=\sqrt{3}-1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}-1\left(x\ge\dfrac{1}{2}\right)\\2x-1=1-\sqrt{3}\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\left(nhận\right)\\x=\dfrac{2-\sqrt{3}}{2}\left(nhận\right)\end{matrix}\right.\)
