a) Ta có: \(P=\left(\frac{2x}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
\(=\left(\frac{2x}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\frac{x+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x+1}{x+1}+\frac{\sqrt{x}}{x+1}\right)\)
\(=\frac{x-1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{x+1}{x+1+\sqrt{x}}\)
\(=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b) Ta có: \(x=\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\)
\(=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\frac{\sqrt{5}+2-\sqrt{5}+2}{\left(\sqrt{5}\right)^2-2^2}\)
\(=\frac{4}{5-4}=\frac{4}{1}=4\)
Thay x=4 vào biểu thức \(P=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\), ta được:
\(P=\frac{\sqrt{4}+1}{4+\sqrt{4}+1}=\frac{2+1}{4+2+1}=\frac{3}{7}\)
Vậy: \(\frac{3}{7}\) là giá trị của biểu thức \(P=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\) tại \(x=\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\)