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