\(\text{Đặt }\left\{{}\begin{matrix}\sqrt{4x^2+5x+1}=a\ge0\\2\sqrt{x^2-x+1}=b\ge0\end{matrix}\right.\Leftrightarrow a^2-b^2=9x-3\\ PT\Leftrightarrow\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\\ \Leftrightarrow a-b=a^2-b^2\\ \Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\a=1-b\end{matrix}\right.\\ \bullet a=b\\ \Leftrightarrow4x^2+5x+1=4x^2-4x+4\\ \Leftrightarrow x=\dfrac{1}{3}\\ \bullet a=1-b\\ \Leftrightarrow\sqrt{4x^2+5x+1}=1-2\sqrt{x^2-x+1}\\ \Leftrightarrow4x^2+5x+1=1+4x^2-4x+4-4\sqrt{x^2-x+1}\\ \Leftrightarrow9x-4=-4\sqrt{x^2-x+1}\\ \Leftrightarrow81x^2-72x+16=16x^2-16x+1\\ \Leftrightarrow65x^2-56x+15=0\left(\text{vô nghiệm}\right)\)
Vậy \(S=\left\{\dfrac{1}{3}\right\}\)
