[( $\frac{1}{√x}$ + $\frac{1}{√y}$ ) × $\frac{2}{√x + √y}$ + $\frac{1}{x}$ + $\frac{1}{y}$] ÷ $\frac{√x³ + y√x + x√y +√y³ }{√x³y +√xy³ }$

Đáp án:

$\begin{array}{l}
\left[ {\left( {\dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt y }}} \right).\dfrac{2}{{\sqrt x  + \sqrt y }} + \dfrac{1}{x} + \dfrac{1}{y}} \right]\\
:\dfrac{{\sqrt {{x^3}}  + y\sqrt x  + x\sqrt y  + \sqrt {{y^3}} }}{{\sqrt {{x^3}y}  + \sqrt {x{y^3}} }}\\
 = \left[ {\dfrac{{\sqrt x  + \sqrt y }}{{\sqrt {xy} }}.\dfrac{2}{{\sqrt x  + \sqrt y }} + \dfrac{{x + y}}{{xy}}} \right]\\
.\dfrac{{\sqrt {xy} \left( {\sqrt x  + \sqrt y } \right)}}{{\sqrt x \left( {x + y} \right) + \sqrt y \left( {x + y} \right)}}\\
 = \left( {\dfrac{2}{{\sqrt {xy} }} + \dfrac{{x + y}}{{xy}}} \right).\dfrac{{\sqrt {xy} \left( {\sqrt x  + \sqrt y } \right)}}{{\left( {\sqrt x  + \sqrt y } \right)\left( {x + y} \right)}}\\
 = \dfrac{{2\sqrt {xy}  + x + y}}{{xy}}.\dfrac{{\sqrt {xy} }}{{x + y}}\\
 = \dfrac{{{{\left( {\sqrt x  + \sqrt y } \right)}^2}}}{{\sqrt {xy} \left( {x + y} \right)}}
\end{array}$