Đáp án:
a. \(\dfrac{{\sqrt {xy} }}{{x - \sqrt {xy} + y}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;y \ge 0;x \ne y\\
P = \left[ {\dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x - \sqrt y }} + \dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{ - \left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right]:\dfrac{{x - 2\sqrt {xy} + y + \sqrt {xy} }}{{\sqrt x + \sqrt y }}\\
= \left[ {\sqrt x + \sqrt y - \dfrac{{x + \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}} \right].\dfrac{{\sqrt x + \sqrt y }}{{x - \sqrt {xy} + y}}\\
= \dfrac{{x + 2\sqrt {xy} + y - x - \sqrt {xy} - y}}{{\sqrt x + \sqrt y }}.\dfrac{{\sqrt x + \sqrt y }}{{x - \sqrt {xy} + y}}\\
= \dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }}.\dfrac{{\sqrt x + \sqrt y }}{{x - \sqrt {xy} + y}}\\
= \dfrac{{\sqrt {xy} }}{{x - \sqrt {xy} + y}}\\
b.P \ge 0\\
\to \dfrac{{\sqrt {xy} }}{{x - \sqrt {xy} + y}} \ge 0\\
\to \left\{ \begin{array}{l}
\sqrt {xy} \ge 0\left( {ld} \right)\forall x \ge 0;y \ge 0;x \ne \\
x - \sqrt {xy} + y > 0\left( {ld} \right)\forall x \ge 0;y \ge 0;x \ne y
\end{array} \right.\\
\to dpcm
\end{array}\)
