Rút gọn biểu thức \(A\).

Điều kiện: \(x \ge 0.\)

\(\begin{array}{l}A = \dfrac{{ - 4x - 9\sqrt x  + 3}}{{x + 3\sqrt x  + 2}} + \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} - \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 2}}\\\,\,\,\,\, = \dfrac{{ - 4x - 9\sqrt x  + 3}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}} + \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} - \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 2}}\\\,\,\,\,\, = \dfrac{{ - 4x - 9\sqrt x  + 3 + \left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right) - \left( {2\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \dfrac{{ - 4x - 9\sqrt x  + 3 + x + \sqrt x  - 2 - 2x - \sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \dfrac{{ - 5x - 9\sqrt x  + 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}} = \dfrac{{ - 5x - 10\sqrt x  + \sqrt x  + 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \dfrac{{ - 5\sqrt x \left( {\sqrt x  + 2} \right) + \left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}} = \dfrac{{\left( {1 - 5\sqrt x } \right)\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}} = \dfrac{{1 - 5\sqrt x }}{{\sqrt x  + 1}}\end{array}\)

Vậy \(A= \dfrac{{1 - 5\sqrt x }}{{\sqrt x  + 1}}\) với \(x \ge 0.\)