a) ĐKXĐ: \(\left\{ \begin{array}{l}x \ne 9\\x \ne 4\\x \ge 0\end{array} \right.\) (*)
\(P = \left( {\frac{{x - 3\sqrt x }}{{x - 9}} - 1} \right):\left( {\frac{{9 - x}}{{x + \sqrt x - 6}} - \frac{{\sqrt x - 3}}{{2 - \sqrt x }} - \frac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right)\)
\[ = \left[ {\frac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} - 1} \right]:\left[ {\frac{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}} + \frac{{\sqrt x - 3}}{{\sqrt x - 2}} - \frac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right]\]
\[ = \left( {\frac{{\sqrt x }}{{\sqrt x + 3}} - 1} \right):\left( {\frac{{3 - \sqrt x }}{{\sqrt x - 2}} + \frac{{\sqrt x - 3}}{{\sqrt x - 2}} - \frac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right)\]
\[ = \frac{{\sqrt x - \sqrt x - 3}}{{\sqrt x + 3}}:\left( {\frac{{3 - \sqrt x + \sqrt x - 3}}{{\sqrt x - 2}} - \frac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right)\]
\[ = \frac{{ - 3}}{{\sqrt x + 3}}:\left( { - \frac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right)\]
\[ = \frac{{ - 3}}{{\sqrt x + 3}}.\frac{{\sqrt x + 3}}{{2 - \sqrt x }}\]
\[ = \frac{3}{{\sqrt x - 2}}\].
b) Ta có \(P < 1 \Leftrightarrow \frac{3}{{\sqrt x - 2}} < 1\).
\( \Leftrightarrow \frac{{3 - \sqrt x + 2}}{{\sqrt x - 2}} < 0\).
\( \Leftrightarrow \frac{{5 - \sqrt x }}{{\sqrt x - 2}} < 0\).
\( \Leftrightarrow \left\{ \begin{array}{l}5 - \sqrt x < 0\\\sqrt x - 2 > 0\end{array} \right.\) hoặc \(\left\{ \begin{array}{l}5 - \sqrt x > 0\\\sqrt x - 2 < 0\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}\sqrt x > 5\\\sqrt x > 2\end{array} \right.\) hoặc \(\left\{ \begin{array}{l}\sqrt x < 5\\\sqrt x < 2\end{array} \right.\)
\( \Leftrightarrow \sqrt x > 5\) hoặc \(\sqrt x < 2\).
⇔ x > 25 hoặc x < 4.
So với điều kiện (*), ta nhận x > 25 hoặc 0 ≤ x < 4.
Vậy x > 25 hoặc 0 ≤ x < 4 thỏa mãn yêu cầu bài toán.