\(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\\ =\left[\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\dfrac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\\ =\dfrac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{4x+8\sqrt{x}}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\\ =\dfrac{4x}{\sqrt{x}-3}\)
