\(a.P=\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right):\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+3}{2\sqrt{x}-x}\right)=\dfrac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{4-x}:\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{8\sqrt{x}+4x}{4-x}.\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right).\sqrt{x}}{\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\dfrac{4x}{\sqrt{x}-3}\) ( x # 4 ; x # 9 ; x > 0 )

\(b.\) \(P=\dfrac{4x}{\sqrt{x}-3}=\dfrac{4\left(x-9\right)+36}{\sqrt{x}-3}=\dfrac{4\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\dfrac{36}{\sqrt{x}-3}=4\sqrt{x}+12+\dfrac{36}{\sqrt{x}-3}=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)

Áp dụng BĐT Cauchy cho các số dương , ta có :

\(4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}\text{≥}2\sqrt{4\left(\sqrt{x}-3\right).\dfrac{36}{\sqrt{x}-3}}=2.2.6=24\) ⇔ \(4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\) ≥ \(24+24=48\)

⇒ \(P_{MIN}=48."="\text{⇔}x=36\)