a) Ta có: \(S=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)
\(=\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{x+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b) Thay \(x=4-2\sqrt{3}\) vào S, ta được:
\(S=\dfrac{4-2\sqrt{3}+\sqrt{3}-1+1}{\sqrt{3}-1-1}=\dfrac{4-\sqrt{3}}{\sqrt{3}-2}\)
\(=\left(\sqrt{3}-4\right)\left(2+\sqrt{3}\right)\)
\(=2\sqrt{3}+3-8-4\sqrt{3}\)
\(=-2\sqrt{3}-5\)
c) Thay \(x=\dfrac{1}{2}\) vào S, ta được:
\(S=\left(\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}+1\right):\left(\dfrac{\sqrt{2}}{2}-1\right)\)
\(=\dfrac{3+\sqrt{2}}{2}:\dfrac{\sqrt{2}-2}{2}\)
\(=\dfrac{3+\sqrt{2}}{\sqrt{2}-2}=\dfrac{\left(3+\sqrt{2}\right)\left(\sqrt{2}+2\right)}{-2}\)
\(=\dfrac{3\sqrt{2}+6+2+2\sqrt{2}}{-2}\)
\(=\dfrac{-5\sqrt{2}-8}{2}\)
