Đáp án:
\[ - 1\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x > 0\\
y > 0\\
x \ne y
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\left( {\dfrac{{\sqrt y }}{{x - \sqrt {xy} }} + \dfrac{{\sqrt x }}{{y - \sqrt {xy} }}} \right):\dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}\\
= \left( {\dfrac{{\sqrt y }}{{\sqrt x \left( {\sqrt x - \sqrt y } \right)}} + \dfrac{{\sqrt x }}{{\sqrt y \left( {\sqrt y - \sqrt x } \right)}}} \right):\dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}\\
= \left( {\dfrac{{\sqrt y }}{{\sqrt x \left( {\sqrt x - \sqrt y } \right)}} - \dfrac{{\sqrt x }}{{\sqrt y \left( {\sqrt x - \sqrt y } \right)}}} \right):\dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}\\
= \dfrac{{\sqrt y .\sqrt y - \sqrt x .\sqrt x }}{{\sqrt x .\sqrt y .\left( {\sqrt x - \sqrt y } \right)}}.\dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }}\\
= \dfrac{{y - x}}{{\left( {\sqrt x - \sqrt y } \right).\left( {\sqrt x + \sqrt y } \right)}}\\
= \dfrac{{y - x}}{{x - y}}\\
= - 1
\end{array}\)
