Đáp án:
$\begin{array}{l}
a > 0;b > 0\\
\dfrac{{a\sqrt a + b\sqrt b }}{{\sqrt a + \sqrt b }} - \sqrt {ab} \\
= \dfrac{{{{\left( {\sqrt a } \right)}^3} + {{\left( {\sqrt b } \right)}^3}}}{{\sqrt a + \sqrt b }} - \sqrt {ab} \\
= \dfrac{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}{{\sqrt a + \sqrt b }} - \sqrt {ab} \\
= a - \sqrt {ab} + b - \sqrt {ab} \\
= a - 2\sqrt {ab} + b\\
= {\left( {\sqrt a - \sqrt b } \right)^2}\\
Vậy\,\dfrac{{a\sqrt a + b\sqrt b }}{{\sqrt a + \sqrt b }} - \sqrt {ab} = {\left( {\sqrt a - \sqrt b } \right)^2}
\end{array}$
