Đáp án:

$\begin{array}{l}
B1)x \ge 0;x \ne 1;x \ne \dfrac{1}{4}\\
B = \left( {\dfrac{{x\sqrt x  + x + \sqrt x }}{{x\sqrt x  - 1}} - \dfrac{{\sqrt x  + 3}}{{1 - \sqrt x }}} \right).\dfrac{{x - 1}}{{2x + \sqrt x  - 1}}\\
 = \left( {\dfrac{{\sqrt x \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} + \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 1}}} \right)\\
.\dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {2\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  + \sqrt x  + 3}}{{\sqrt x  - 1}}.\dfrac{{\sqrt x  - 1}}{{2\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + 3}}{{2\sqrt x  - 1}}\\
B < 0\\
 \Rightarrow \dfrac{{2\sqrt x  + 3}}{{2\sqrt x  - 1}} < 0\\
 \Rightarrow 2\sqrt x  - 1 < 0\left( {do:2\sqrt x  + 3 > 0} \right)\\
 \Rightarrow \sqrt x  < \dfrac{1}{2}\\
 \Rightarrow x < \dfrac{1}{4}\\
\text{Vậy}\,0 \le x < \dfrac{1}{4}\\
B2)\\
a > 0;a \ne 1\\
C = \left( {\dfrac{6}{{a - 1}} + \dfrac{{10 - 2\sqrt a }}{{a\sqrt a  - a - \sqrt a  + 1}}} \right).\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \left( {\dfrac{6}{{a - 1}} + \dfrac{{10 - 2\sqrt a }}{{\left( {a - 1} \right)\left( {\sqrt a  - 1} \right)}}} \right).\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \dfrac{{6\left( {\sqrt a  - 1} \right) + 10 - 2\sqrt a }}{{\left( {\sqrt a  - 1} \right)\left( {a - 1} \right)}}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \dfrac{{6\sqrt a  - 6 + 10 - 2\sqrt a }}{{\sqrt a  + 1}}.\dfrac{1}{{4\sqrt a }}\\
 = \dfrac{{4\sqrt a  + 4}}{{\sqrt a  + 1}}.\dfrac{1}{{4\sqrt a }}\\
 = \dfrac{1}{{\sqrt a }}\\
D = C.\left( {a - \sqrt a  + 1} \right)\\
 = \dfrac{{a - \sqrt a  + 1}}{{\sqrt a }} = \sqrt a  - 1 + \dfrac{1}{{\sqrt a }}\\
 = \sqrt a  + \dfrac{1}{{\sqrt a }} - 1\\
Theo\,Co - si:\sqrt a  + \dfrac{1}{{\sqrt a }} \ge 2\sqrt {\sqrt a .\dfrac{1}{{\sqrt a }}}  = 2\\
 \Rightarrow \sqrt a  + \dfrac{1}{{\sqrt a }} - 1 \ge 2 - 1 = 1\\
 \Rightarrow D \ge 1\\
\text{Vậy}\,D \ge 1
\end{array}$