Rút gọn $D = \left( {\dfrac{{\sqrt x  + \sqrt y }}{{1 - \sqrt {xy} }} - \dfrac{{\sqrt x  - \sqrt y }}{{1 + \sqrt {xy} }}} \right):\left( {\dfrac{{y + xy}}{{1 - xy}}} \right)$ với \(x \ge 0;\,\,y \ge 0;\,\,xy \ne 1\) và $M = \left( {\dfrac{{x\sqrt x }}{{\sqrt x  + 1}} + \dfrac{{{x^2}}}{{x\sqrt x  + x}}} \right)\left( {2 - \dfrac{1}{{\sqrt x }}} \right)$ với \(x > 0\) ta được

$D = \left( {\dfrac{{\sqrt x  + \sqrt y }}{{1 - \sqrt {xy} }} - \dfrac{{\sqrt x  - \sqrt y }}{{1 + \sqrt {xy} }}} \right):\left( {\dfrac{{y + xy}}{{1 - xy}}} \right)$$ = \dfrac{{\left( {\sqrt x  + \sqrt y } \right)\left( {1 + \sqrt {xy} } \right) - \left( {\sqrt x  - \sqrt y } \right)\left( {1 - \sqrt {xy} } \right)}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}.\dfrac{{1 - xy}}{{y + xy}}$$ = \dfrac{{\sqrt x  + \sqrt y  + x\sqrt y  + y\sqrt x  - \left( {\sqrt x  - \sqrt y  - x\sqrt y  + y\sqrt x } \right)}}{{1 - xy}}.\dfrac{{1 - xy}}{{y + xy}}$$ = \dfrac{{\sqrt x  + \sqrt y  + x\sqrt y  + y\sqrt x  - \sqrt x  + \sqrt y  + x\sqrt y  - y\sqrt x }}{{y + xy}}$$ = \dfrac{{2\sqrt y  + 2x\sqrt y }}{{y + xy}} = \dfrac{{2\sqrt y \left( {x + 1} \right)}}{{y\left( {x + 1} \right)}} = \dfrac{2}{{\sqrt y }}$

Vậy \(D = \dfrac{2}{{\sqrt y }}\)  với \(x \ge 0;\,\,y \ge 0;\,\,xy \ne 1\)

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

Vậy \(M = 2x - \sqrt x \)  với \(x > 0.\)