Đáp án: $\dfrac{{{{\left( {x + y} \right)}^2}}}{{2xy}}$
Giải thích các bước giải:
$\begin{array}{l}
C = \dfrac{1}{{{{\left( {\dfrac{{\sqrt x + \sqrt y }}{{\sqrt {x + y} }} - \dfrac{{\sqrt {x + y} }}{{\sqrt x + \sqrt y }}} \right)}^2}}} - \dfrac{{x + y}}{{2\sqrt x .\sqrt y }} - \dfrac{{\sqrt {{{\left( {x + y} \right)}^4}} }}{{4xy}}\\
= \dfrac{1}{{{{\left( {\dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2} - \left( {x + y} \right)}}{{\left( {\sqrt {x + y} } \right).\left( {\sqrt x + \sqrt y } \right)}}} \right)}^2}}} - \dfrac{{x + y}}{{2\sqrt x .\sqrt y }} - \dfrac{{{{\left( {x + y} \right)}^2}}}{{4xy}}\\
= \dfrac{{\left( {x + y} \right).{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{{{\left( {x + 2\sqrt {xy} + y - x - y} \right)}^2}}} - \dfrac{{\left( {x + y} \right).2\sqrt {xy} - {{\left( {x + y} \right)}^2}}}{{4xy}}\\
= \dfrac{{\left( {x + y} \right)\left( {x + 2\sqrt {xy} + y} \right) - 2\sqrt {xy} \left( {x + y} \right) + {{\left( {x + y} \right)}^2}}}{{4xy}}\\
= \dfrac{{2\sqrt {xy} \left( {x + y} \right) + {{\left( {x + y} \right)}^2} - 2\sqrt {xy} \left( {x + y} \right) + {{\left( {x + y} \right)}^2}}}{{4xy}}\\
= \dfrac{{2{{\left( {x + y} \right)}^2}}}{{4xy}}\\
= \dfrac{{{{\left( {x + y} \right)}^2}}}{{2xy}}
\end{array}$
