\(a.P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)=\dfrac{8\sqrt{x}+4x}{4-x}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\dfrac{4x}{\sqrt{x}-3}\) ( x # 4 ; x > 0 ; x # 9 )
\(a.P=-1\text{ }\text{⇒}\dfrac{4x}{\sqrt{x}-3}=-1\)
\(\text{⇔}4x=3-\sqrt{x}\text{⇔}4x+\sqrt{x}-3=0\text{⇔}4x+4\sqrt{x}-3\sqrt{x}-3=0\text{⇔}4\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=0\text{⇔}\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\text{⇔}x=\dfrac{9}{16}\left(TM\right)\) Vậy ,...
