Đáp án:
$\begin{array}{l}
Dkxd:a > 0;b > 0;a \ne b\\
B = \left( {\dfrac{{\sqrt a }}{{\sqrt {ab} - b}} + \dfrac{{\sqrt b }}{{\sqrt {ab} - a}}} \right).\left( {a\sqrt a - b\sqrt a } \right)\\
= \left( {\dfrac{{\sqrt a }}{{\sqrt b \left( {\sqrt a - \sqrt b } \right)}} + \dfrac{{\sqrt b }}{{\sqrt a \left( {\sqrt b - \sqrt a } \right)}}} \right).\\
\sqrt a .\left( {a - b} \right)\\
= \dfrac{{\sqrt a .\sqrt a - \sqrt b .\sqrt b }}{{\sqrt a .\sqrt b \left( {\sqrt a - \sqrt b } \right)}}.\sqrt a .\left( {\sqrt a - \sqrt b } \right).\left( {\sqrt a + \sqrt b } \right)\\
= \dfrac{{a - b}}{{\sqrt b }}.\left( {\sqrt a + \sqrt b } \right)\\
= \dfrac{{a\sqrt a + a\sqrt b - b\sqrt a - b\sqrt b }}{{\sqrt b }}
\end{array}$
