Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne \dfrac{1}{4}
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
P = \left( {1 - \dfrac{{\sqrt x + 4x}}{{4x - 1}}} \right):\left( {\dfrac{{1 - 2x}}{{1 - 4x}} + \dfrac{{2\sqrt x }}{{2\sqrt x - 1}} - 1} \right)\\
= \dfrac{{\left( {4x - 1} \right) - \left( {\sqrt x + 4x} \right)}}{{4x - 1}}:\left( {\dfrac{{1 - 2x}}{{\left( {1 - 2\sqrt x } \right)\left( {1 + 2\sqrt x } \right)}} - \dfrac{{2\sqrt x }}{{1 - 2\sqrt x }} - 1} \right)\\
= \dfrac{{ - 1 - \sqrt x }}{{4x - 1}}:\dfrac{{\left( {1 - 2x} \right) - 2\sqrt x \left( {1 + 2\sqrt x } \right) - \left( {1 - 2\sqrt x } \right)\left( {1 + 2\sqrt x } \right)}}{{\left( {1 - 2\sqrt x } \right)\left( {1 + 2\sqrt x } \right)}}\\
= \dfrac{{ - \left( {\sqrt x + 1} \right)}}{{4x - 1}}:\dfrac{{1 - 2x - 2\sqrt x - 4x - \left( {1 - 4x} \right)}}{{\left( {1 - 2\sqrt x } \right)\left( {1 + 2\sqrt x } \right)}}\\
= \dfrac{{ - \left( {\sqrt x + 1} \right)}}{{4x - 1}}:\dfrac{{ - 2x - 2\sqrt x }}{{1 - 4x}}\\
= \dfrac{{ - \left( {\sqrt x + 1} \right)}}{{4x - 1}}:\dfrac{{ - 2\sqrt x \left( {\sqrt x + 1} \right)}}{{1 - 4x}}\\
= \dfrac{{ - \left( {\sqrt x + 1} \right)}}{{4x - 1}}.\dfrac{{1 - 4x}}{{ - 2\sqrt x .\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 1}}{{2\sqrt x }}
\end{array}\)
