$\begin{array}{l} A = \left( {\dfrac{{2 - x}}{{2 - x}} - \dfrac{{2 - x}}{{x + 2}} - \dfrac{{4{x^2}}}{{{x^2} - 4}}} \right):\dfrac{{x - 3}}{{2x - {x^2}}}\\ A = \left( {\dfrac{{2 + x}}{{2 - x}} - \dfrac{{2 - x}}{{x + 2}} + \dfrac{{4{x^2}}}{{4 - {x^2}}}} \right).\dfrac{{2x - {x^2}}}{{x - 3}}\\ A = \dfrac{{{{\left( {2 + x} \right)}^2} - {{\left( {2 - x} \right)}^2} + 4{x^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\dfrac{{x\left( {2 - x} \right)}}{{x - 3}}\\ A = \dfrac{{{x^2} + 4x + 4 - {x^2} + 4x - 4 + 4{x^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\dfrac{{x\left( {2 - x} \right)}}{{x - 3}}\\ A = \dfrac{{\left( {4{x^2} + 8x} \right)x}}{{\left( {2 + x} \right)\left( {x - 3} \right)}} = \dfrac{{4{x^2}\left( {x + 2} \right)}}{{\left( {2 + x} \right)\left( {x - 3} \right)}} = \dfrac{{4{x^2}}}{{x - 3}}\\ D = \dfrac{1}{{2\sqrt x - 2}} - \dfrac{1}{{2\sqrt x + 2}} + \dfrac{{\sqrt x }}{{1 - x}}\\ D = \dfrac{1}{{2\left( {\sqrt x - 1} \right)}} - \dfrac{1}{{2\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ D = \dfrac{{\left( {\sqrt x + 1} \right) - \left( {\sqrt x - 1} \right) - 2\sqrt x }}{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ D = \dfrac{{2 - 2\sqrt x }}{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{{2\left( {1 - \sqrt x } \right)}}{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = - \dfrac{1}{{\sqrt x + 1}} \end{array}$
