c) Có: $\frac{1}{P}$ < $\frac{1}{6}$ Đkxđ: x ≥ 0; x $\neq$ 9
⇔ $\frac{1}{\frac{\sqrt{x} - 2}{\sqrt{x} - 3}}$ < $\frac{1}{6}$
⇔ $\frac{\sqrt{x} - 3}{\sqrt{x} - 2}$ < $\frac{1}{6}$
⇔ $\frac{\sqrt{x} - 3}{\sqrt{x} - 2}$ - $\frac{1}{6}$ < 0
⇔ $\frac{6.(\sqrt{x} - 3) - 1.(\sqrt{x} - 2)}{6.(\sqrt{x} - 2)}$ < 0
⇔ $\frac{6\sqrt{x} - 18 - \sqrt{x} + 2}{6\sqrt{x} - 12}$ < 0
⇔ $\frac{5\sqrt{x} -16}{6\sqrt{x} - 12}$ < 0
⇔ \(\left[ \begin{array}{l}\left \{ {{5\sqrt{x} - 16 > 0} \atop {6\sqrt{x} - 12 < 0}} \right.\\\left \{ {{5\sqrt{x} - 16 < 0} \atop {6\sqrt{x} - 12 > 0}} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\left \{ {{\sqrt{x} > \frac{16}{5}} \atop {\sqrt{x} < 2}} \right.\\\left \{ {{\sqrt{x} < \frac{16}{5}} \atop {\sqrt{x} > 2}} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\left \{ {{x > \frac{256}{25}} \atop {x < 4}} \right. (Vô lí)\\\left \{ {{x < \frac{256}{25}} \atop {x > 4}} \right. \end{array} \right.\)
⇔ 4 < x < $\frac{256}{25}$
Vậy 4 < x < $\frac{256}{25}$ thì $\frac{1}{P}$ < $\frac{1}{6}$
