Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x - \sqrt x \ne 0\\
x + \sqrt x - 2 \ne 0\\
x - 2\sqrt x + 1 \ne 0\\
\sqrt x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \left( {\sqrt x - 1} \right) \ne 0\\
\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) \ne 0\\
{\left( {\sqrt x - 1} \right)^2} \ne 0\\
\sqrt x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
P = \left( {\dfrac{1}{{x - \sqrt x }} + \dfrac{{\sqrt x + 2}}{{x + \sqrt x - 2}}} \right):\dfrac{{\sqrt x + 1}}{{x - 2\sqrt x + 1}}\\
= \left( {\dfrac{1}{{\sqrt x .\left( {\sqrt x - 1} \right)}} + \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}} \right):\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \left( {\dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b,\\
P > - \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x }} > - \dfrac{1}{2}\\
\Leftrightarrow \sqrt x - 1 > - \dfrac{1}{2}\sqrt x \\
\Leftrightarrow \dfrac{3}{2}\sqrt x > 1\\
\Leftrightarrow \sqrt x > \dfrac{2}{3}\\
\Leftrightarrow x > \dfrac{4}{9}\\
\Rightarrow \left\{ \begin{array}{l}
x > \dfrac{4}{9}\\
x \ne 1
\end{array} \right.
\end{array}\)
