Đáp án:
$\begin{array}{l}
a)Dkxd:x,y \ge 0;x\# y\\
A = \left( {\dfrac{{x - y}}{{\sqrt x - \sqrt y }} + \dfrac{{x\sqrt x - y\sqrt y }}{{y - x}}} \right):{\left( {\sqrt x - \sqrt y } \right)^2}\\
+ \dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }}\\
= \left( {\sqrt x + \sqrt y - \dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right)\\
:{\left( {\sqrt x - \sqrt y } \right)^2} + \dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }}\\
= \left( {\sqrt x + \sqrt y - \dfrac{{x + \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}} \right):{\left( {\sqrt x - \sqrt y } \right)^2} + \dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }}\\
= \dfrac{{x + 2\sqrt {xy} + y - x - \sqrt {xy} - y}}{{\sqrt x + \sqrt y }}.\dfrac{1}{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}} + \dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }}\\
= \dfrac{{\sqrt {xy} }}{{\sqrt x + y}}.\dfrac{1}{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}} + \dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }}\\
= \left( {\dfrac{1}{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}} + 1} \right).\dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }} \ge 0\\
Do:x;y \ge 0\\
Vay\,A \ge 0\\
Bt2)\\
B = \left( {\dfrac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}}} \right) - \dfrac{{2a + \sqrt a }}{{\sqrt a }} - 1\\
= \dfrac{{\sqrt a \left( {a\sqrt a + 1} \right)}}{{a - \sqrt a + 1}} - \dfrac{{\sqrt a \left( {2\sqrt a + 1} \right)}}{{\sqrt a }} - 1\\
= \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{a - \sqrt a + 1}} - \left( {2\sqrt a + 1} \right) - 1\\
= \sqrt a \left( {\sqrt a + 1} \right) - 2\sqrt a - 1 + 1\\
= a + \sqrt a - 2\sqrt a \\
= a - \sqrt a
\end{array}$
