Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
b \ge 0\\
\sqrt a - \sqrt b \ne 0\\
a - b \ne 0\\
\sqrt {ab} - \left( {a + b} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a \ge 0\\
b \ge 0\\
a \ne b
\end{array} \right.\\
b,\\
A = \left( {\dfrac{{\sqrt a + \sqrt b }}{{\sqrt a - \sqrt b }} - \dfrac{{4\sqrt {ab} }}{{a - b}}} \right).\left( {\dfrac{{a\sqrt a + b\sqrt b }}{{\sqrt {ab} - \left( {a + b} \right)}}} \right)\\
= \left( {\dfrac{{\sqrt a + \sqrt b }}{{\sqrt a - \sqrt b }} - \dfrac{{4\sqrt {ab} }}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}} \right).\dfrac{{{{\sqrt a }^3} + {{\sqrt b }^3}}}{{ - \left( {a - \sqrt {ab} + b} \right)}}\\
= \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2} - 4\sqrt {ab} }}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\dfrac{{\left( {\sqrt a + \sqrt b } \right)\left( {{{\sqrt a }^2} - \sqrt a .\sqrt b + {{\sqrt b }^2}} \right)}}{{ - \left( {a - \sqrt {ab} + b} \right)}}\\
= \dfrac{{\left( {a + 2\sqrt {ab} + b} \right) - 4\sqrt {ab} }}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\dfrac{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}{{ - \left( {a - \sqrt {ab} + b} \right)}}\\
= \dfrac{{a - 2\sqrt {ab} + b}}{{\left( {\sqrt a - \sqrt b } \right).\left( {\sqrt a + \sqrt b } \right)}}.\dfrac{{\sqrt a + \sqrt b }}{{ - 1}}\\
= \dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\dfrac{{\sqrt a + \sqrt b }}{{ - 1}}\\
= - \left( {\sqrt a - \sqrt b } \right)\\
= \sqrt b - \sqrt a
\end{array}\)
