a. ĐK: x ≠ 2; x ≠ 3; x ≥ 0

\(A = \frac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \frac{{\sqrt x + 3}}{{\sqrt x - 2}} - \frac{{2\sqrt x + 1}}{{3 - \sqrt x }} = \frac{{2\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \frac{{\sqrt x + 3}}{{\sqrt x  - 2}} + \frac{{2\sqrt x + 1}}{{\sqrt x - 3}}\)

\(A = \frac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\)

\(A = \frac{{2\sqrt x - 9 - \left( {x - 9} \right) + 2x - 4\sqrt x + \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}} = \frac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\)

\(A = \frac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}} = \frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}} = \frac{{\sqrt x + 1}}{{\sqrt x + 3}}\).

b. ĐK: x ≥ 0

A < 1 \( \Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x + 3}} - 1 < 0\)

\( \Leftrightarrow \frac{{\sqrt x + 1 - \sqrt x - 3}}{{\sqrt x + 3}} < 0 \Leftrightarrow \frac{{ - 2}}{{\sqrt x + 3}} < 0\)

Do –2 < 0

Để \(\frac{{ - 2}}{{\sqrt x + 3}} < 0 \Leftrightarrow \sqrt x + 3 > 0\) (luôn đúng do x ≥ 0)

⇒ ∀x ∈ ℝ thì A < 1.