ĐKXĐ: \(x\ge-8\)
\(\Leftrightarrow3\sqrt{3}\left(x^2+4x+2\right)=\sqrt{x+8}\) (với \(x^2+4x+2\ge0\))
\(\Rightarrow27\left(x^2+4x+2\right)^2=x+8\)
\(\Leftrightarrow27x^4+216x^3+540x^2+431x+100=0\)
\(\Leftrightarrow\left(3x^2+11x+4\right)\left(9x^2+39x+25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x^2+11x+4=0\\9x^2+39x+25=0\end{matrix}\right.\)