\(\dfrac{A}{B}=\dfrac{2\sqrt{x}-2}{3}\left(ĐK:x>0\right)\\ < =>\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{2\sqrt{x}-2}{3}\\ < =>\dfrac{\sqrt{x}+2}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{2\sqrt{x}-2}{3}\\ < =>\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}-2}{3}\)
\(< =>\sqrt{x}\left(2\sqrt{x}-2\right)=3\left(\sqrt{x}+1\right)\\ < =>2x-2\sqrt{x}=3\sqrt{x}+3\\ < =>2x-5\sqrt{x}-3=0\\ < =>\left(\sqrt{x}-3\right)\left(2\sqrt{x}+1\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\2\sqrt{x}+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=9\left(TMDK\right)\\\sqrt{x}=-\dfrac{1}{2}< 0\left(KTMDK\right)\end{matrix}\right.< =>x=9\)
