ĐKXĐ: \(x\ge0;x\ne4\)
\(P=\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x}+4}{x-4}\)
\(=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(x+1\right)+2\left(x+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+1}{\sqrt{x}-2}\)
Khi \(x=9+4\sqrt{5}\)
Ta có: \(4+4\sqrt{5}+5=2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=\left(2+\sqrt{5}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{5}\)
\(\Rightarrow P=\dfrac{\left(2+\sqrt{5}\right)^2+1}{2+\sqrt{5}-2}=\dfrac{9+4\sqrt{5}+1}{\sqrt{5}}=\dfrac{10+4\sqrt{5}}{\sqrt{5}}=4+2\sqrt{5}\)
Vậy \(P=4+2\sqrt{5}\) khi \(x=9+4\sqrt{5}\).