Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
2 + \sqrt x \ne 0\\
4 - x \ne 0\\
x - 2\sqrt x \ne 0\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right.\\
b,\\
P = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{4 - x}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \dfrac{2}{{\sqrt x }}} \right)\\
= \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \dfrac{2}{{\sqrt x }}} \right)\\
= \dfrac{{4\sqrt x \left( {2 - \sqrt x } \right) + 8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{\sqrt x - 1 - 2.\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{8\sqrt x - 4x + 8x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{\sqrt x - 1 - 2\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4x + 8\sqrt x }}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{ - \sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4\sqrt x \left( {\sqrt x + 2} \right)}}{{ - \left( {\sqrt x - 2} \right).\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x .\left( {\sqrt x - 2} \right)}}{{ - \left( {\sqrt x - 3} \right)}}\,\,\,\,\,\,\,\,\left( {x \ne 9} \right)\\
= \dfrac{{4x}}{{\sqrt x - 3}}\\
c,\\
P > 1 \Leftrightarrow \dfrac{{4x}}{{\sqrt x - 3}} > 1\\
\Leftrightarrow \dfrac{{4x}}{{\sqrt x - 3}} - 1 > 0\\
\Leftrightarrow \dfrac{{4x - \left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}} > 0\\
\Leftrightarrow \dfrac{{4x - \sqrt x + 3}}{{\sqrt x - 3}} > 0\\
4x - \sqrt x + 3 = {\left( {2\sqrt x } \right)^2} - 2.2\sqrt x .\dfrac{1}{4} + {\left( {\dfrac{1}{4}} \right)^2} + \dfrac{{47}}{{16}} = {\left( {2\sqrt x - \dfrac{1}{4}} \right)^2} + \dfrac{{47}}{{16}} > 0\\
\Rightarrow \sqrt x - 3 > 0 \Leftrightarrow x > 9
\end{array}\)
