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