\((2x-3)(x-\dfrac{1}{2} )= 0 \)

\(⇒\left[\begin{matrix}2x-3=0\\ x-\dfrac{1}{2}=0\end{matrix}\right. ⇔\left[\begin{matrix}x=\dfrac{3}{2}\\ x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x \in \)\(\left\{\dfrac{3}{2};\dfrac{1}{2}\right\}\)

\(\dfrac{x-4}{\sqrt{x}+2} + \dfrac{x+2\sqrt{x}}{\sqrt{x}} ( x>0) \)

\(A\) \(= \dfrac{\sqrt{x}(x-4)}{\sqrt{x}(\sqrt{x}+2)} + \dfrac{(x+2\sqrt{x})(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)} \)

\(A\) \(= \dfrac{x\sqrt{x}-4\sqrt{x}+x\sqrt{x}+2x+2x+4\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)} \)

\(A\)\(= \dfrac{2x\sqrt{x}+4x}{\sqrt{x}(\sqrt{x}+2)} =\dfrac{2x(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)} =2\sqrt{x}\)

Vậy \(A=\) \(2\sqrt{x}\)