Đáp án:

$\begin{array}{l}
B1)\\
a)Dkxd:x \ge 2\\
\sqrt {4x - 8}  - \sqrt {9x - 18}  + \sqrt {\dfrac{{x - 2}}{{25}}}  =  - 3\\
 \Leftrightarrow \sqrt {4\left( {x - 2} \right)}  - \sqrt {9\left( {x - 2} \right)}  + \dfrac{{\sqrt {x - 2} }}{5} =  - 3\\
 \Leftrightarrow 2\sqrt {x - 2}  - 3\sqrt {x - 2}  + \dfrac{1}{5}\sqrt {x - 2}  =  - 3\\
 \Rightarrow \dfrac{{ - 4}}{5}\sqrt {x - 2}  =  - 3\\
 \Rightarrow \sqrt {x - 2}  = \dfrac{{15}}{4}\\
 \Rightarrow x - 2 = \dfrac{{225}}{{16}}\\
 \Rightarrow x = \dfrac{{257}}{{16}}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{{257}}{{16}}\\
b)Dkxd:x \ge \dfrac{{ - 4}}{5}\\
x - \sqrt {5x + 4}  = 2\\
 \Rightarrow x - 2 = \sqrt {5x + 4} \\
 \Rightarrow {x^2} - 4x + 4 = 5x + 4\\
 \Rightarrow {x^2} - 9x = 0\\
 \Rightarrow x\left( {x - 9} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = 9\left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 0;x = 9\\
B2)\\
Dkxd:x \ge 0;x \ne 4\\
a)x = 11 + 6\sqrt 2 \left( {tmdk} \right)\\
 = 9 + 2.3.\sqrt 2  + 2\\
 = {\left( {3 + \sqrt 2 } \right)^2}\\
 \Rightarrow \sqrt x  = 3 + \sqrt 2 \\
A = \dfrac{{x + 3}}{{\sqrt x  - 2}} = \dfrac{{11 + 6\sqrt 2  + 3}}{{3 + \sqrt 2  - 2}}\\
 = \dfrac{{14 + 6\sqrt 2 }}{{1 + \sqrt 2 }}\\
 = \dfrac{{\left( {14 + 6\sqrt 2 } \right)\left( {\sqrt 2  - 1} \right)}}{{2 - 1}}\\
 = 14\sqrt 2  - 14 + 12 - 6\sqrt 2 \\
 = 8\sqrt 2  - 2\\
b)B = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}} + \dfrac{{5\sqrt x  - 2}}{{x - 4}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right) + 5\sqrt x  - 2}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2 + 5\sqrt x  - 2}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  - 2}}\\
c)C = \dfrac{A}{B}\\
 = \dfrac{{x + 3}}{{\sqrt x  - 2}}:\dfrac{{\sqrt x }}{{\sqrt x  - 2}}\\
 = \dfrac{{x + 3}}{{\sqrt x }} = \sqrt x  + \dfrac{3}{{\sqrt x }}\\
Theo\,\text{Cô} - si:\\
\sqrt x  + \dfrac{3}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{3}{{\sqrt x }}}  = 2\sqrt 3 \\
 \Rightarrow C \ge 2\sqrt 3 \\
 \Rightarrow GTNN:C = 2\sqrt 3 \\
Khi:\sqrt x  = \dfrac{3}{{\sqrt x }} \Rightarrow x = 3 \Rightarrow x = \sqrt 3 
\end{array}$