Đáp án:

$\begin{array}{l}
1)a)Dkxd:x \ge 0;x \ne 4;x \ne 9\\
A = \left( {\dfrac{{x - 2\sqrt x }}{{x - 4}} - 1} \right)\\
:\left( {\dfrac{{4 - x}}{{x - \sqrt x  - 6}} - \dfrac{{\sqrt x  - 2}}{{3 - \sqrt x }} - \dfrac{{\sqrt x  - 3}}{{\sqrt x  + 2}}} \right)\\
 = \left( {\dfrac{{\sqrt x \left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} - 1} \right)\\
:\dfrac{{4 - x + \left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right) - \left( {\sqrt x  - 3} \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  + 2}} - 1} \right).\dfrac{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 2} \right)}}{{4 - x + x - 4 - \left( {x - 6\sqrt x  + 9} \right)}}\\
 = \dfrac{{\sqrt x  - \sqrt x  - 2}}{{\sqrt x  + 2}}.\dfrac{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 2} \right)}}{{ - {{\left( {\sqrt x  - 3} \right)}^2}}}\\
 = \dfrac{{ - 2}}{1}.\dfrac{{\left( {\sqrt x  - 3} \right)}}{{ - {{\left( {\sqrt x  - 3} \right)}^2}}}\\
 = \dfrac{2}{{\sqrt x  - 3}}\\
b)A \in Z\\
 \Rightarrow \dfrac{2}{{\sqrt x  - 3}} \in Z\\
 \Rightarrow \left( {\sqrt x  - 3} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ {1;2;4;5} \right\}\\
 \Rightarrow x \in \left\{ {1;4;16;25} \right\}\\
Do:x \ne 4;x \ne 9\\
 \Rightarrow x \in \left\{ {1;16;25} \right\}\\
B2)\\
a)A = \dfrac{{\sqrt {17 + 12\sqrt 2 }  - 5\sqrt {17 - 12\sqrt 2 } }}{{\sqrt 2  - 1}}\\
 = \dfrac{{\sqrt {9 + 2.3.2\sqrt 2  + 8}  - 5.\sqrt {9 - 2.3.2\sqrt 2  + 8} }}{{\sqrt 2  - 1}}\\
 = \dfrac{{\sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}}  - 5\sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} }}{{\sqrt 2  - 1}}\\
 = \dfrac{{3 + 2\sqrt 2  - 5\left( {3 - 2\sqrt 2 } \right)}}{{\sqrt 2  - 1}}\\
 = \dfrac{{12\sqrt 2  - 12}}{{\sqrt 2  - 1}}\\
 = 12\\
b)\cos a = \dfrac{1}{3}\\
Do:0 < a < {90^0} \Rightarrow \sin a > 0\\
{\sin ^2}a + {\cos ^2}a = 1\\
 \Rightarrow {\sin ^2}a = 1 - {\cos ^2}a = \dfrac{8}{9}\\
 \Rightarrow \sin a = \dfrac{{2\sqrt 2 }}{3}\\
B = \dfrac{{{\mathop{\rm sina}\nolimits}  - 3cosa}}{{\sin a + 2\cos a}} = \dfrac{{\dfrac{{2\sqrt 2 }}{3} - 3.\dfrac{1}{3}}}{{\dfrac{{2\sqrt 2 }}{3} + 2.\dfrac{1}{3}}} = \dfrac{{2\sqrt 2  - 3}}{{2\sqrt 2  + 2}}\\
 = \dfrac{{\left( {2\sqrt 2  - 3} \right).\left( {\sqrt 2  - 1} \right)}}{{2.\left( {2 - 1} \right)}}\\
 = \dfrac{{4 - 2\sqrt 2  - 3\sqrt 2  + 3}}{2}\\
 = \dfrac{{7 - 5\sqrt 2 }}{2}
\end{array}$