Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
B = \left( {\dfrac{{\sqrt x + \sqrt y }}{{2\sqrt x - 2\sqrt y }} - \dfrac{{2\sqrt {xy} }}{{x - y}}} \right).\dfrac{{2\sqrt x }}{{\sqrt x - \sqrt y }}\\
= \left( {\dfrac{{\sqrt x + \sqrt y }}{{2.\left( {\sqrt x - \sqrt y } \right)}} - \dfrac{{2\sqrt {xy} }}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right).\dfrac{{2\sqrt x }}{{\sqrt x - \sqrt y }}\\
= \dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2} - 4.\sqrt {xy} }}{{2\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}.\dfrac{{2\sqrt x }}{{\sqrt x - \sqrt y }}\\
= \dfrac{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}}{{2.\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}.\dfrac{{2\sqrt x }}{{\sqrt x - \sqrt y }}\\
= \dfrac{{\sqrt x }}{{\sqrt x + \sqrt y }}\\
B - 1 = \dfrac{{\sqrt x }}{{\sqrt x + \sqrt y }} - 1 = \dfrac{{\sqrt x - \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x + \sqrt y }} = \dfrac{{ - \sqrt y }}{{\sqrt x + \sqrt y }} < 0,\,\,\,\forall x,y > 0\\
\Rightarrow B < 1
\end{array}\)
