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