Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
P = \left( {\dfrac{{2 + \sqrt x }}{{2 - \sqrt x }} - \dfrac{{2 - \sqrt x }}{{2 + \sqrt x }} - \dfrac{{4x}}{{x - 4}}} \right):\dfrac{{\sqrt x - 3}}{{2\sqrt x - x}}\\
= \left( {\dfrac{{2 + \sqrt x }}{{2 - \sqrt x }} - \dfrac{{2 - \sqrt x }}{{2 + \sqrt x }} - \dfrac{{4x}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right):\dfrac{{\sqrt x - 3}}{{\sqrt x .\left( {2 - \sqrt x } \right)}}\\
= \dfrac{{{{\left( {2 + \sqrt x } \right)}^2} - {{\left( {2 - \sqrt x } \right)}^2} + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{\sqrt x - 3}}{{\sqrt x .\left( {2 - \sqrt x } \right)}}\\
= \dfrac{{\left( {x + 4\sqrt x + 4} \right) - \left( {x - 4\sqrt x + 4} \right) + 4x}}{{\left( {2 - \sqrt x } \right).\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x .\left( {2 - \sqrt x } \right)}}{{\sqrt x - 3}}\\
= \dfrac{{4x + 8\sqrt x }}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x .\left( {2 - \sqrt x } \right)}}{{\sqrt x - 3}}\\
= \dfrac{{4\sqrt x \left( {\sqrt x + 2} \right)}}{{2 + \sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
= \dfrac{{4\sqrt x .\sqrt x }}{{\sqrt x - 3}}\\
= \dfrac{{4x}}{{\sqrt x - 3}}\\
b,\\
\left| P \right| = 1 \Leftrightarrow \left[ \begin{array}{l}
P = 1\\
P = - 1
\end{array} \right.\\
TH1:\,\,\,P = 1 \Leftrightarrow \dfrac{{4x}}{{\sqrt x - 3}} = 1\\
\Leftrightarrow 4x = \sqrt x - 3\\
\Leftrightarrow 4x - \sqrt x + 3 = 0\\
\Leftrightarrow {\left( {2\sqrt x } \right)^2} - 2.2\sqrt x .\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{47}}{{16}} = 0\\
\Leftrightarrow {\left( {2\sqrt x - \dfrac{1}{4}} \right)^2} + \dfrac{{47}}{{16}} = 0\,\,\,\,\,\left( {vn} \right)\\
TH2:\,\,\,P = - 1 \Leftrightarrow \dfrac{{4x}}{{\sqrt x - 3}} = - 1\\
\Leftrightarrow 4x = - \sqrt x + 3\\
\Leftrightarrow 4x + \sqrt x - 3 = 0\\
\Leftrightarrow \left( {\sqrt x + 1} \right)\left( {4\sqrt x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = - 1\\
\sqrt x = \dfrac{3}{4}
\end{array} \right. \Leftrightarrow \sqrt x = \dfrac{3}{4} \Leftrightarrow x = \dfrac{9}{{16}}
\end{array}\)
