\(\sqrt{3x^2+6x+16}+\sqrt{x^2+2x}=2\sqrt{x^2+2x+4}\) \(\Leftrightarrow\sqrt{3\left(x^2+2x\right)+16}+\sqrt{x^2+2x}=2\sqrt{x^2+2x+4}\) Đặt \(t=x^2+2x\), \(\left(t\ge0\right)\) phương trình trở thành: \(\sqrt{3t+16}+\sqrt{t}=2\sqrt{t+4}\) \(3t+16+t+2\sqrt{t\left(3t+16\right)}=4\left(t+4\right)\) \(\Leftrightarrow2\sqrt{t\left(3t+16\right)}=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=0\\3t+16=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\) \(\Leftrightarrow t=0\). Với \(t=0\Leftrightarrow x^2+2x=0\) \(\Leftrightarrow x\left(x+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\). Vậy x = 0 hoặc x = -2.