Đáp án:
$\begin{array}{l}
A = \dfrac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2}\\
= \dfrac{{{{\left( {\sqrt x } \right)}^3} + {{\left( {\sqrt y } \right)}^3}}}{{\sqrt x + \sqrt y }} - x + 2\sqrt {xy} - y\\
= \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {x - \sqrt {xy} + y} \right)}}{{\sqrt x + \sqrt y }} - x + 2\sqrt {xy} - y\\
= x - \sqrt {xy} + y - x + 2\sqrt {xy} - y\\
= \sqrt {xy} \\
B = \dfrac{{\sqrt a + a\sqrt b - \sqrt b - b\sqrt a }}{{ab - 1}}\\
= \dfrac{{\sqrt a \left( {1 + \sqrt {ab} } \right) - \sqrt b \left( {1 + \sqrt {ab} } \right)}}{{\left( {\sqrt {ab} + 1} \right)\left( {\sqrt {ab} - 1} \right)}}\\
= \dfrac{{\left( {\sqrt {ab} + 1} \right)\left( {\sqrt a - \sqrt b } \right)}}{{\left( {\sqrt {ab} + 1} \right)\left( {\sqrt {ab} - 1} \right)}}\\
= \dfrac{{\sqrt a - \sqrt b }}{{\sqrt {ab} - 1}}
\end{array}$
