Đáp án:
$\begin{array}{l}
Dkxd:x > 0;x \ne 1;x \ne 4\\
a)P = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x }}} \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\\
= \dfrac{{\sqrt x - \left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{x - 1 - x + 4}}\\
= \dfrac{1}{{\sqrt x }}.\dfrac{{\sqrt x - 2}}{3}\\
= \dfrac{{\sqrt x - 2}}{{3\sqrt x }}\\
b)P > \dfrac{1}{6}\\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{{3\sqrt x }} - \dfrac{1}{6} > 0\\
\Leftrightarrow \dfrac{{2\sqrt x - 4 - \sqrt x }}{{6\sqrt x }} > 0\\
\Leftrightarrow \sqrt x - 4 > 0\\
\Leftrightarrow \sqrt x > 4\\
\Leftrightarrow x > 16\\
Vậy\,x > 16\\
c)P < \dfrac{1}{4}\\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{{3\sqrt x }} - \dfrac{1}{4} < 0\\
\Leftrightarrow \dfrac{{4\sqrt x - 8 - 3\sqrt x }}{{12\sqrt x }} < 0\\
\Leftrightarrow \sqrt x - 8 < 0\\
\Leftrightarrow \sqrt x < 8\\
\Leftrightarrow x < 64\\
Vậy\,0 < x < 64;x \ne 1;x \ne 4
\end{array}$
