Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne \dfrac{1}{9}\\
P = \left( {\dfrac{{\sqrt x - 1}}{{3\sqrt x - 1}} - \dfrac{1}{{3\sqrt x + 1}} + \dfrac{{8\sqrt x }}{{9x - 1}}} \right):\left( {1 - \dfrac{{3\sqrt x - 2}}{{3\sqrt x + 1}}} \right)\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 1} \right) - 3\sqrt x + 1 + 8\sqrt x }}{{\left( {3\sqrt x - 1} \right)\left( {3\sqrt x + 1} \right)}}:\dfrac{{3\sqrt x + 1 - 3\sqrt x + 2}}{{3\sqrt x + 1}}\\
= \dfrac{{3x + \sqrt x - 3\sqrt x - 1 + 5\sqrt x + 1}}{{\left( {3\sqrt x - 1} \right)\left( {3\sqrt x + 1} \right)}}.\dfrac{{3\sqrt x + 1}}{3}\\
= \dfrac{{3x + 3\sqrt x }}{{3\sqrt x - 1}}.\dfrac{1}{3}\\
= \dfrac{{x + \sqrt x }}{{3\sqrt x - 1}}\\
b)P = \dfrac{6}{5}\\
\Rightarrow \dfrac{{x + \sqrt x }}{{3\sqrt x - 1}} = \dfrac{6}{5}\\
\Rightarrow 5x + 5\sqrt x = 18\sqrt x - 6\\
\Rightarrow 5x - 13\sqrt x + 6 = 0\\
\Rightarrow 5x - 3\sqrt x - 10\sqrt x + 6 = 0\\
\Rightarrow \left( {5\sqrt x - 3} \right)\left( {\sqrt x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = \dfrac{3}{5}\\
\sqrt x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{9}{{25}}\left( {tmdk} \right)\\
x = 4\left( {tmdk} \right)
\end{array} \right.
\end{array}$
