Rút gọn $A + B$ ta được:

Điều kiện: $x \ge 0,\,\,x \ne 9.$

$\begin{array}{l}A + B = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\sqrt x \left( {\sqrt x  - 3} \right) + 2\sqrt x \left( {\sqrt x  + 3} \right) - 3x - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{x - 3\sqrt x  + 2x + 6\sqrt x  - 3x - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{3\sqrt x  - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{3\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{3}{{\sqrt x  + 3}}\end{array}$

Vậy $A + B = \dfrac{3}{{\sqrt x  + 3}}$ (với $x \ge 0,\,\,x \ne 9$).