Đáp án:
$\begin{array}{l}
Dkxd:a \ge 0;b \ge 0;a\# b\\
A = \dfrac{{a\sqrt a + b\sqrt b }}{{\sqrt a + \sqrt b }}:\left( {a - b} \right) + \dfrac{{2\sqrt b }}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}{{\left( {\sqrt a + \sqrt b } \right)\left( {a - b} \right)}} + \dfrac{{2\sqrt b }}{{\sqrt a + \sqrt b }}\\
= \dfrac{{a - \sqrt {ab} + b}}{{a - b}} + \dfrac{{2\sqrt b }}{{\sqrt a + \sqrt b }}\\
= \dfrac{{a - \sqrt {ab} + b + 2\sqrt b \left( {\sqrt a - \sqrt b } \right)}}{{\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}\\
= \dfrac{{a - \sqrt {ab} + b + 2\sqrt {ab} - 2b}}{{\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}\\
= \dfrac{{a + \sqrt {ab} - b}}{{a - b}}\\
= \dfrac{{a - b}}{{a - b}} + \dfrac{{\sqrt {ab} }}{{a - b}}\\
= 1 + \dfrac{{\sqrt {ab} }}{{a - b}}
\end{array}$
