\(x+\sqrt{4x^2-4x+1}=2\left(đk:x\le2\right)\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=2-x\)
\(\Leftrightarrow\left|2x-1\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2-x\left(2\ge x\ge\dfrac{1}{2}\right)\\2x-1=x-2\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(x+\sqrt{4x^2-4x+1}=2\)
\(\Leftrightarrow x+\sqrt{\left(2x\right)^2-2.2x.1+1^2}=2\)
\(\Leftrightarrow x+\sqrt{\left(2x-1\right)^2}=2\)
\(\Leftrightarrow\left|2x-1\right|=2-x\)
\(\Leftrightarrow2x-1=2-x\) hoặc \(2x-1=x-2\)
\(\Leftrightarrow3x=3\) \(\Leftrightarrow x=-1\)
\(\Leftrightarrow x=1\)
Vậy S = \(\left\{1;-1\right\}\)
