Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1\\
P = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{\sqrt x + 1}} + \dfrac{{\sqrt x }}{{1 - x}}} \right)\\
:\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} + \dfrac{{1 - \sqrt x }}{{\sqrt x + 1}}} \right)\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} + \sqrt x \left( {\sqrt x - 1} \right) - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
:\dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1 + x - \sqrt x - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1}}\\
= \dfrac{{2x + 1}}{1}.\dfrac{1}{{4\sqrt x }}\\
= \dfrac{{2x + 1}}{{4\sqrt x }}\\
b)x = \dfrac{{2 - \sqrt 3 }}{2}\left( {tmdk} \right)\\
= \dfrac{{4 - 2\sqrt 3 }}{4}\\
= \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{4}\\
\Rightarrow \sqrt x = \dfrac{{\sqrt 3 - 1}}{2}\\
\Rightarrow P = \dfrac{{2x + 1}}{{4\sqrt x }}\\
= \dfrac{{2.\dfrac{{2 - \sqrt 3 }}{2} + 1}}{{4.\dfrac{{\sqrt 3 - 1}}{2}}}\\
= \dfrac{{2 - \sqrt 3 + 1}}{{2\sqrt 3 - 2}}\\
= \dfrac{{3 - \sqrt 3 }}{{2\left( {\sqrt 3 - 1} \right)}} = \dfrac{{\sqrt 3 }}{2}\\
c)P - \dfrac{1}{2}\\
= \dfrac{{2x + 1}}{{4\sqrt x }} - \dfrac{1}{2}\\
= \dfrac{{2x + 1 - 2\sqrt x }}{{4\sqrt x }}\\
= \dfrac{{2\left( {x - \sqrt x } \right) + 1}}{{4\sqrt x }}\\
= \dfrac{{2.\left( {x - \sqrt x + \dfrac{1}{4}} \right) - 2.\dfrac{1}{4} + 1}}{{4\sqrt x }}\\
= \dfrac{{2{{\left( {\sqrt x - \dfrac{1}{2}} \right)}^2} + \dfrac{1}{2}}}{{4\sqrt x }} > 0\forall x > 0;x \ne 1\\
\Rightarrow P - \dfrac{1}{2} > 0\\
\Rightarrow P > \dfrac{1}{2}
\end{array}$
