ĐK : \(x\ne0vsx\ge-3\)
\(PT\Leftrightarrow4x^2+x+3-4x\sqrt{x+3}=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2=0\)
\(\Leftrightarrow2x=\sqrt{x+3}\) ( ĐK : \(x\ge0\) )
\(\Leftrightarrow4x^2-x-3=0\)
\(\Delta=1+48=49>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{1+7}{8}=1\left(N\right)\\x_2=\frac{1-7}{8}=-\frac{3}{4}\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)