ĐKXĐ: \(x>0;x\ne4\)
\(P=\left(\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\frac{4x+2\sqrt{x}-4}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{2\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}-\frac{\sqrt{x}+3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(=\left(\frac{x+4\sqrt{x}+4+2\sqrt{x}-x+4x+2\sqrt{x}-4}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(=\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-3\right)}=\frac{4x}{\sqrt{x}-3}\)
\(P>0\Rightarrow\frac{4x}{\sqrt{x}-3}>0\Rightarrow\sqrt{x}-3>0\Rightarrow x>9\)
\(P=-1\Rightarrow\frac{4x}{\sqrt{x}-3}=-1\Rightarrow4x=-\sqrt{x}+3\)
\(\Rightarrow4x+\sqrt{x}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(l\right)\\\sqrt{x}=\frac{3}{4}\end{matrix}\right.\) \(\Rightarrow x=\frac{9}{16}\)
