Đáp án:
$\begin{array}{l}
a)Dk{\rm{xd}}:x > 0;x \ne 1;x \ne 4\\
P = \left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} + \dfrac{{2\sqrt x - 1}}{{\sqrt x - x}}} \right):\dfrac{{\sqrt x - 1}}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x .\sqrt x - 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}.\left( {\sqrt x - 2} \right)\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}\left( {\sqrt x - 2} \right)}}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
b)P < 0\\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{{\sqrt x }} < 0\\
\Leftrightarrow \sqrt x - 2 < 0\left( {do:\sqrt x > 0} \right)\\
\Leftrightarrow \sqrt x < 2\\
\Leftrightarrow x < 4\\
Vậy\,0 < x < 4;x \ne 1\\
c)P = - 3\sqrt x \\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{{\sqrt x }} = - 3\sqrt x \\
\Leftrightarrow \sqrt x - 2 = - 3x\\
\Leftrightarrow 3x + \sqrt x - 2 = 0\\
\Leftrightarrow 3x + 3\sqrt x - 2\sqrt x - 2 = 0\\
\Leftrightarrow \left( {\sqrt x + 1} \right)\left( {3\sqrt x - 2} \right) = 0\\
\Leftrightarrow \sqrt x = \dfrac{2}{3}\\
\Leftrightarrow x = \dfrac{4}{9}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{4}{9}\\
d)x = 6 - 4\sqrt 2 \left( {tmdk} \right)\\
= 4 - 2.2.\sqrt 2 + 2\\
= {\left( {2 - \sqrt 2 } \right)^2}\\
\Leftrightarrow \sqrt x = 2 - \sqrt 2 \\
\Leftrightarrow P = \dfrac{{\sqrt x - 2}}{{\sqrt x }} = \dfrac{{2 - \sqrt 2 - 2}}{{2 - \sqrt 2 }}\\
= \dfrac{{ - \sqrt 2 }}{{2 - \sqrt 2 }}\\
= \dfrac{1}{{1 - \sqrt 2 }}\\
= \dfrac{{1 + \sqrt 2 }}{{1 - 2}}\\
= - 1 - \sqrt 2
\end{array}$
