$\begin{array}{l} C = \left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{3}{{x - 1}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{x - \sqrt x - 3}}{{x + 2\sqrt x + 1}}} \right)\\ = \dfrac{{\sqrt x - 1 + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) - x + \sqrt x + 3}}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\ = \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{x - 1 - x + \sqrt x + 3}}\\ = \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\ = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\ D = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} + \dfrac{{3\sqrt x }}{{1 - \sqrt x }}} \right):\dfrac{1}{{\sqrt x + 1}}\\ = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2} - 3\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\left( {\sqrt x + 1} \right)\\ = \dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1 - 3\sqrt x }}{{\sqrt x - 1}}\\ = \dfrac{{\sqrt x }}{{\sqrt x - 1}} \end{array}$
