Đáp án:

$\begin{array}{l}
B1)\\
a)Dkxd:x \ge 0;x \ne 9\\
M = \left( {\dfrac{{x + \sqrt x  + 10}}{{x - 9}} - \dfrac{1}{{\sqrt x  - 3}}} \right):\dfrac{1}{{\sqrt x  - 3}}\\
 = \dfrac{{x + \sqrt x  + 10 - \sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\left( {\sqrt x  - 3} \right)\\
 = \dfrac{{x + 7}}{{\sqrt x  + 3}}\\
b)x = 25\left( {tmdk} \right)\\
 \Rightarrow \sqrt x  = 5\\
 \Rightarrow M = \dfrac{{x + 7}}{{\sqrt x  + 3}} = \dfrac{{25 + 7}}{{5 + 3}} = 4\\
c)x \ge 0;x \ne 9\\
M < \sqrt x  + 1\\
 \Rightarrow \dfrac{{x + 7}}{{\sqrt x  + 3}} < \sqrt x  + 1\\
 \Rightarrow x + 7 < \left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 3} \right)\\
 \Rightarrow x + 7 < x + 4\sqrt x  + 3\\
 \Rightarrow 4\sqrt x  > 4\\
 \Rightarrow \sqrt x  > 1\\
 \Rightarrow x > 1\\
Vay\,x > 1;x \ne 9\\
B2)\\
a)Dkxd:x \ge 0;x \ne 9\\
x = 1\\
 \Rightarrow \sqrt x  = 1\\
 \Rightarrow B = \dfrac{1}{{\sqrt x  - 3}} = \dfrac{1}{{1 - 3}} =  - \dfrac{1}{2}\\
b)S = A - B\\
 = \dfrac{{x + 3}}{{x - 9}} + \dfrac{2}{{\sqrt x  + 3}} - \dfrac{1}{{\sqrt x  - 3}}\\
 = \dfrac{{x + 3 + 2\left( {\sqrt x  - 3} \right) - \left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{x + 3 + 2\sqrt x  - 6 - \sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{x + \sqrt x  - 6}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{\sqrt x  - 2}}{{\sqrt x  - 3}}\\
c)S = \dfrac{{\sqrt x  + 1}}{{\sqrt x  + 2}} = \dfrac{{\sqrt x  - 2}}{{\sqrt x  - 3}}\\
 \Rightarrow \left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 3} \right) = \left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)\\
 \Rightarrow x - 2\sqrt x  - 3 = x - 4\\
 \Rightarrow 2\sqrt x  = 1\\
 \Rightarrow \sqrt x  = \dfrac{1}{2}\\
 \Rightarrow x = \dfrac{1}{4}\left( {tmdk} \right)
\end{array}$