\(\sqrt{x}+\sqrt{3x}+\sqrt{4x}=\dfrac{2\sqrt{3}}{\sqrt{3}-1}\)
\(\Rightarrow\sqrt{x}+\sqrt{3x}+\sqrt{4}\sqrt{x}=\dfrac{2\sqrt{3}\cdot\left(\sqrt{3}+1\right)}{2}\)
\(\Leftrightarrow\sqrt{x}+\sqrt{3x}+2\sqrt{x}=\sqrt{3}\cdot\left(\sqrt{3}+1\right)\)
\(\Leftrightarrow3\sqrt{x}+\sqrt{3x}=3+\sqrt{3}\)
\(\Leftrightarrow\left(3\sqrt{x}+\sqrt{3x}\right)^2=\left(3+\sqrt{3}\right)^2\)
\(\Leftrightarrow9x+6\sqrt{3x^2}+3x=9+6\sqrt{3}+3\)
\(\Leftrightarrow9x+6\sqrt{3}\cdot\left|x\right|+3x=12+6\sqrt{3}\)
\(\Leftrightarrow12x+6\sqrt{3}\cdot\left|x\right|=12+6\sqrt{3}\)
\(\Rightarrow\left[{}\begin{matrix}12x+6\sqrt{3}x=12+6\sqrt{3}\left(đk:x\ge0\right)\\12x+6\sqrt{3}\cdot\left(-x\right)=12+6\sqrt{3}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(đk:x\ge0\right)\\x=7+4\sqrt{3}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
