Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne \dfrac{1}{9}\\
A = \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) - \left( {3\sqrt x - 1} \right) + 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 - 2\sqrt x - 1 - 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}\\
= \dfrac{{3x + 3\sqrt x }}{{3\sqrt x - 1}}.\dfrac{1}{3}\\
= \dfrac{{x + \sqrt x }}{{3\sqrt x - 1}}\\
b)x = 6 + 2\sqrt 5 \\
= {\left( {\sqrt 5 + 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 5 + 1\\
\Rightarrow A = \dfrac{{x + \sqrt x }}{{3\sqrt x - 1}}\\
= \dfrac{{6 + 2\sqrt 5 + \sqrt 5 + 1}}{{3\left( {\sqrt 5 + 1} \right) - 1}}\\
= \dfrac{{7 + 3\sqrt 5 }}{{2 + 3\sqrt 5 }}\\
= \dfrac{{31 + 15\sqrt 5 }}{{41}}\\
c)A = \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 - 10\sqrt x - 3\sqrt x + 6 = 0\\
\Rightarrow \left( {\sqrt x - 2} \right)\left( {5\sqrt x - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = \dfrac{3}{5}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tmdk} \right)\\
x = \dfrac{9}{{25}}\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,x = 4;x = \dfrac{9}{{25}}\\
d)A < 1\\
\Rightarrow \dfrac{{x + \sqrt x }}{{3\sqrt x - 1}} < 1\\
\Rightarrow \dfrac{{x + \sqrt x - 3\sqrt x + 1}}{{3\sqrt x - 1}} < 0\\
\Rightarrow \dfrac{{x - 2\sqrt x + 1}}{{3\sqrt x - 1}} < 0\\
\Rightarrow \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{3\sqrt x - 1}} < 0\\
\Rightarrow 3\sqrt x - 1 < 0\left( {do:{{\left( {\sqrt x - 1} \right)}^2} \ge 0} \right)\\
\Rightarrow \sqrt x < \dfrac{1}{3}\\
\Rightarrow x < \dfrac{1}{9}\\
Vậy\,0 \le x < \dfrac{1}{9}
\end{array}$