Đáp án:

$\begin{array}{l}
1)a)Dkxd:x \ge 0;x \ne 9\\
B = \left( {\dfrac{1}{{\sqrt x  + 3}} + \dfrac{3}{{x\sqrt x  - 9\sqrt x }}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x  + 3}} - \dfrac{{3\sqrt x  - 3}}{{x + 3\sqrt x }}} \right)\\
 = \left( {\dfrac{1}{{\sqrt x  + 3}} + \dfrac{3}{{\sqrt x \left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}} \right)\\
:\dfrac{{\sqrt x .\sqrt x  - 3\sqrt x  + 3}}{{\sqrt x .\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 3} \right) + 3}}{{\sqrt x \left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  + 3} \right)}}{{x - 3\sqrt x  + 3}}\\
 = \dfrac{{x - 3\sqrt x  + 3}}{{\sqrt x  - 3}}.\dfrac{1}{{x - 3\sqrt x  + 3}}\\
 = \dfrac{1}{{\sqrt x  - 3}}\\
b)B > 1\\
 \Rightarrow \dfrac{1}{{\sqrt x  - 3}} - 1 > 0\\
 \Rightarrow \dfrac{{1 - \sqrt x  + 3}}{{\sqrt x  - 3}} > 0\\
 \Rightarrow \dfrac{{\sqrt x  - 4}}{{\sqrt x  - 3}} < 0\\
 \Rightarrow 3 < \sqrt x  < 4\\
 \Rightarrow 9 < x < 16\\
2)a)Dkxd:x \ge 0\\
C = \dfrac{{x + 2}}{{x\sqrt x  + 1}} + \dfrac{{\sqrt x  - 1}}{{x + \sqrt x  + 1}} - \dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{{x + 2 + \left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right) - x - \sqrt x  - 1}}{{\left( {\sqrt x  + 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{x + 2 + x - 1 - x - \sqrt x  - 1}}{{\left( {\sqrt x  + 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{x - \sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{x\sqrt x  + 1}}\\
b)??\\
3)a)Dkxd:x \ge 0;x \ne 1\\
D = \left( {\dfrac{{2\sqrt x  + x}}{{x\sqrt x  - 1}} - \dfrac{1}{{\sqrt x  - 1}}} \right):\dfrac{{x - 1}}{{x + \sqrt x  + 1}}\\
 = \dfrac{{2\sqrt x  + x - \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\dfrac{{x + \sqrt x  + 1}}{{x - 1}}\\
 = \dfrac{{2\sqrt x  + x - x - \sqrt x  - 1}}{{\sqrt x  - 1}}.\dfrac{1}{{x - 1}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  - 1}}.\dfrac{1}{{x - 1}}\\
 = \dfrac{1}{{x - 1}}\\
b)D <  - \dfrac{1}{2}\\
 \Rightarrow \dfrac{1}{{x - 1}} <  - \dfrac{1}{2}\\
 \Rightarrow \dfrac{1}{{x - 1}} + \dfrac{1}{2} < 0\\
 \Rightarrow \dfrac{{2 + x - 1}}{{2\left( {x - 1} \right)}} < 0\\
 \Rightarrow \dfrac{{x + 1}}{{2\left( {x - 1} \right)}} < 0\\
 \Rightarrow x - 1 < 0\left( {do:x \ge 0} \right)\\
 \Rightarrow x < 1\\
Vay\,0 \le x < 1
\end{array}$

(B2 câu b ko thể luôn có giá trị ko âm với mọi x được.)