a: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{1}{x-1}\)
b: Thay \(x=5+2\sqrt{3}\) vào A, ta được:
\(A=\dfrac{1}{4+2\sqrt{3}}=\dfrac{4-2\sqrt{3}}{4}\)
\(\Leftrightarrow\sqrt{A}=\dfrac{\sqrt{3}-1}{2}\)
a)\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\left(\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}.\dfrac{1}{x-1}\)
b)\(x=5+2\sqrt{3}\Rightarrow A=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}=\dfrac{1}{3+2\sqrt{3}+1}\)
\(\Leftrightarrow A=\dfrac{1}{\left(\sqrt{3}+1\right)^2}\Rightarrow\sqrt{A}=\dfrac{1}{\sqrt{3}+1}\)
