ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
\(2x^2+4x+3=3\sqrt{\left(x^2+x+1\right)\left(2x+1\right)}\)
\(\Leftrightarrow2\left(x^2+x+1\right)+\left(2x+1\right)-3\sqrt{\left(x^2+x+1\right)\left(2x+1\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{2x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+b^2-3ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+1}=\sqrt{2x+1}\\2\sqrt{x^2+x+1}=\sqrt{2x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=2x+1\\4\left(x^2+x+1\right)=2x+1\end{matrix}\right.\)
\(\Leftrightarrow...\)
