a) Ta có: \(\left(\dfrac{1}{2-\sqrt{3}}-\dfrac{3}{\sqrt{7}-2}\right):\dfrac{2}{\sqrt{7}+\sqrt{3}}\)
\(=\left(2+\sqrt{3}-\sqrt{7}-2\right):\dfrac{\left(\sqrt{7}-\sqrt{3}\right)}{2}\)
\(=\dfrac{-\left(\sqrt{7}-\sqrt{3}\right)}{1}\cdot\dfrac{2}{\sqrt{7}-\sqrt{3}}\)
=-2
b) Ta có: \(\left(\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-1\right):\left(\sqrt{x}-x\right)+\dfrac{1}{x}\)
\(=\left(-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-1\right)\cdot\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{x}\)
\(=\left(-\sqrt{x}-1\right)\cdot\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{x}\)
\(=\dfrac{x+\sqrt{x}}{x\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1}{x\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}-1}{x\left(\sqrt{x}-1\right)}\)
