ĐKXĐ: \(x\ge-2\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)=3\sqrt{x+2}\)
Đặt \(\sqrt{x+2}=t\ge0\)
\(\Rightarrow t^2\left(t^2+2\right)=3t\)
\(\Leftrightarrow t\left(t^3+2t-3\right)=0\)
\(\Leftrightarrow t\left(t-1\right)\left(t^2+t+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=1\end{matrix}\right.\) \(\Rightarrow x=\)