b) ĐK : \(x\ge\dfrac{1}{2}\)
\(\sqrt{x^2+4x}=2x-1\)
\(\Leftrightarrow x^2+4x=4x^2-4x+1\)
\(\Leftrightarrow3x^2-8x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{13}}{3}\left(tm\right)\\x=\dfrac{4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)


