Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;y \ge 0;x \ne y\\
Q = \dfrac{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + \sqrt {xy} }}{{\sqrt x + \sqrt y }}:\left( {\dfrac{{x - y}}{{\sqrt x - \sqrt y }} + \dfrac{{x\sqrt x - y\sqrt y }}{{y - x}}} \right)\\
= \dfrac{{x - \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}\\
:\left( {\sqrt x + \sqrt y + \dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{\left( {\sqrt y - \sqrt x } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right)\\
= \dfrac{{x - \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}:\left( {\sqrt x + \sqrt y - \dfrac{{x + \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}} \right)\\
= \dfrac{{x - \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}:\dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2} - x - \sqrt {xy} - y}}{{\sqrt x + \sqrt y }}\\
= \dfrac{{x - \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}.\dfrac{{\sqrt x + \sqrt y }}{{x + 2\sqrt {xy} + y - x - \sqrt {xy} - y}}\\
= \dfrac{{x - \sqrt {xy} + y}}{{\sqrt {xy} }}\\
b)Q - 1\\
= \dfrac{{x - \sqrt {xy} + y}}{{\sqrt {xy} }} - 1\\
= \dfrac{{x - \sqrt {xy} + y - \sqrt {xy} }}{{\sqrt {xy} }}\\
= \dfrac{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}}{{\sqrt {xy} }} > 0\left( {do:\sqrt x \ne \sqrt y } \right)\\
\Leftrightarrow Q > 1
\end{array}$
