Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = \left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x - 1}}} \right):\left( {\dfrac{2}{x} - \dfrac{{2 - x}}{{x\sqrt x + x}}} \right)\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\left( {\dfrac{2}{x} + \dfrac{{x - 2}}{{x\left( {\sqrt x + 1} \right)}}} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{2\left( {\sqrt x + 1} \right) + x - 2}}{{x.\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{x + 2\sqrt x }}{{x\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{x\left( {\sqrt x + 1} \right)}}{{x + 2\sqrt x }}\\
= \dfrac{x}{{\sqrt x - 1}}\\
a,\\
x = \dfrac{{968}}{{23 - 3\sqrt 5 }} = \dfrac{{1936}}{{46 - 2.3\sqrt 5 }} = \dfrac{{{{44}^2}}}{{45 - 2.3\sqrt 5 + 1}} = \dfrac{{{{44}^2}}}{{{{\left( {3\sqrt 5 - 1} \right)}^2}}}\\
\Rightarrow \sqrt x = \dfrac{{44}}{{3\sqrt 5 - 1}}\\
P = \dfrac{x}{{\sqrt x - 1}} = \dfrac{{\dfrac{{968}}{{23 - 3\sqrt 5 }}}}{{\dfrac{{44}}{{3\sqrt 5 - 1}} - 1}}\\
= \dfrac{{968}}{{23 - 3\sqrt 5 }}:\dfrac{{44 - \left( {3\sqrt 5 - 1} \right)}}{{3\sqrt 5 - 1}}\\
= \dfrac{{968}}{{23 - 3\sqrt 5 }}:\dfrac{{45 - 3\sqrt 5 }}{{3\sqrt 5 - 1}}\\
= \dfrac{{968\left( {3\sqrt 5 - 1} \right)}}{{\left( {23 - 3\sqrt 5 } \right).\left( {45 - 3\sqrt 5 } \right)}}\\
= \dfrac{{968.\left( {3\sqrt 5 - 1} \right)}}{{1080 - 204\sqrt 5 }}\\
= \dfrac{{242\left( {3\sqrt 5 - 1} \right)}}{{270 - 51\sqrt 5 }}\\
b,\\
P > 2 \Leftrightarrow \dfrac{x}{{\sqrt x - 1}} > 2\\
\Leftrightarrow \dfrac{x}{{\sqrt x - 1}} - 2 > 0\\
\Leftrightarrow \dfrac{{x - 2.\left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \dfrac{{x - 2\sqrt x + 2}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \dfrac{{{{\left( {\sqrt x - 1} \right)}^2} + 1}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \sqrt x - 1 > 0\\
\Leftrightarrow x > 1
\end{array}\)
