\(a.E=\left(\dfrac{3+\sqrt{x}}{x-1}+\dfrac{3}{\sqrt{x}+1}\right):\dfrac{4}{x+\sqrt{x}}=\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{4}=\dfrac{x}{\sqrt{x}-1}\) ( x ≥ 0 ; x # 1 )
\(b.x=\dfrac{9}{4}\left(TMĐKXĐ\right)\) ⇒ \(\sqrt{x}=\dfrac{3}{2}\) , ta có :