a: \(P=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4x+8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(=\frac{4x}{\sqrt{x}-3}\)
b: P=-1
=>\(\frac{4x}{\sqrt{x}-3}=-1\)
=>\(4x=-\sqrt{x}+3\)
=>\(4x+\sqrt{x}-3=0\)
=>\(4x+4\sqrt{x}-3\sqrt{x}-3=0\)
=>\(\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\)
=>\(4\sqrt{x}-3=0\)
=>\(\sqrt{x}=\frac34\)
=>x=9/16(nhận)
