đkxđ: x≥0; x≠1
\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\dfrac{\sqrt{x}}{1-x}\right)=\left[\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}+4\sqrt{x}\right]\cdot\left(-\dfrac{x-1}{\sqrt{x}}\right)=-\dfrac{\left(\sqrt{x}+1-\sqrt{x}+1\right)\left(\sqrt{x}+1+\sqrt{x}-1\right)+4\sqrt{x}\left(x-1\right)}{x-1}\cdot\dfrac{x-1}{\sqrt{x}}=-\dfrac{4\sqrt{x}+4\sqrt{x}\left(x-1\right)}{\sqrt{x}}=-\dfrac{4\sqrt{x}\left(1+x-1\right)}{\sqrt{x}}=-4x\)
Ta có: \(x=\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{5}}=\dfrac{\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)}{3-5}=\dfrac{3-\sqrt{15}}{-2}\)
=> \(P=-4\cdot\dfrac{3-\sqrt{15}}{-2}=2\left(3-\sqrt{15}\right)=6-2\sqrt{15}\)
