Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
b \ge 0\\
a \ne b
\end{array} \right.\\
P = \left( {\dfrac{1}{{\sqrt a + \sqrt b }} + \dfrac{{3\sqrt {ab} }}{{a\sqrt a + b\sqrt b }}} \right).\\
\left[ {\left( {\dfrac{1}{{\sqrt a - \sqrt b }} - \dfrac{{3\sqrt {ab} }}{{a\sqrt a - b\sqrt b }}} \right):\dfrac{{a - b}}{{a + \sqrt {ab} + b}}} \right]\\
= \dfrac{{a - \sqrt {ab} + b + 3\sqrt {ab} }}{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}\\
.\dfrac{{a + \sqrt {ab} + b - 3\sqrt {ab} }}{{\left( {\sqrt a - \sqrt b } \right)\left( {a + \sqrt {ab} + b} \right)}}\\
.\dfrac{{a + \sqrt {ab} + b}}{{\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}\\
= \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}.\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{\left( {\sqrt a - \sqrt b } \right)}}\\
.\dfrac{1}{{\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}\\
= \dfrac{1}{{a - \sqrt {ab} + b}}\\
b)Dkxd:a \ge 0;b \ge 0;a \ne b\\
\left\{ \begin{array}{l}
a = 16\left( {tmdk} \right)\\
b = 4\left( {tmdk} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\sqrt a = 4\\
\sqrt b = 2
\end{array} \right.\\
P = \dfrac{1}{{a - \sqrt {ab} + b}} = \dfrac{1}{{16 - 4.2 + 4}} = \dfrac{1}{{12}}
\end{array}$
