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