Đáp án:

c. \(0 < x < \dfrac{4}{{25}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.DK:x > 0;x \ne \dfrac{1}{4}\\
A = \left( {\dfrac{{ - 4x + \sqrt x  - 1 + 4x}}{{1 - 4x}}} \right):\left[ {\dfrac{{1 + 2x + 2\sqrt x \left( {2\sqrt x  + 1} \right) - 1 + 4x}}{{\left( {2\sqrt x  + 1} \right)\left( {1 - 2\sqrt x } \right)}}} \right]\\
 = \dfrac{{\sqrt x  - 1}}{{1 - 4x}}:\dfrac{{1 + 2x + 4x + 2\sqrt x  - 1 + 4x}}{{\left( {2\sqrt x  + 1} \right)\left( {1 - 2\sqrt x } \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{1 - 4x}}:\dfrac{{10x + 2\sqrt x }}{{1 - 4x}}\\
 = \dfrac{{\sqrt x  - 1}}{{10x + 2\sqrt x }}\\
b.A > {A^2}\\
 \to {A^2} - A < 0\\
 \to A\left( {A - 1} \right) < 0\\
 \to \left\{ \begin{array}{l}
A < 1\\
A > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{\sqrt x  - 1}}{{10x + 2\sqrt x }} < 1\\
\dfrac{{\sqrt x  - 1}}{{10x + 2\sqrt x }} > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{\sqrt x  - 1 - 10x - 2\sqrt x }}{{10x + 2\sqrt x }} < 0\\
\dfrac{{\sqrt x  - 1}}{{10x + 2\sqrt x }} > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{ - 10x - \sqrt x  - 1}}{{10x + 2\sqrt x }} < 0\\
\dfrac{{\sqrt x  - 1}}{{10x + 2\sqrt x }} > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
10x + 2\sqrt x  > 0\left( {do: - 10x - \sqrt x  - 1 < 0\forall x > 0} \right)\\
\sqrt x  - 1 > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\sqrt x  > 0\\
\sqrt x  <  - \dfrac{1}{5}\left( l \right)
\end{array} \right.\\
x > 1
\end{array} \right.\\
 \to x > 1\\
c.\left| A \right| > \dfrac{1}{4}\\
 \to \left| {\dfrac{{\sqrt x  - 1}}{{10x + 2\sqrt x }}} \right| > \dfrac{1}{4}\\
 \to \left[ \begin{array}{l}
\dfrac{{\sqrt x  - 1}}{{10x + 2\sqrt x }} > \dfrac{1}{4}\\
\dfrac{{\sqrt x  - 1}}{{10x + 2\sqrt x }} <  - \dfrac{1}{4}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\dfrac{{4\sqrt x  - 4 - 10x - 2\sqrt x }}{{4\left( {10x + 2\sqrt x } \right)}} > 0\\
\dfrac{{4\sqrt x  - 4 + 10x + 2\sqrt x }}{{4\left( {10x + 2\sqrt x } \right)}} < 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\dfrac{{ - 10x + 2\sqrt x  - 4}}{{4\left( {10x + 2\sqrt x } \right)}} > 0\\
\dfrac{{10x + 6\sqrt x  - 4}}{{4\left( {10x + 2\sqrt x } \right)}} < 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
10x + 2\sqrt x  < 0\left( { - 10x + 2\sqrt x  - 4 < 0\forall x > 0} \right)\\
\dfrac{{\left( {5\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}}{{4\left( {10x + 2\sqrt x } \right)}} < 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
 - \dfrac{1}{5} < \sqrt x  < 0\left( {vô lý} \right)\\
\dfrac{{5\sqrt x  - 2}}{{\sqrt x \left( {5\sqrt x  + 2} \right)}} < 0\left( {do:\sqrt x  + 1 > 0\forall x > 0} \right)
\end{array} \right.\\
 \to \dfrac{{5\sqrt x  - 2}}{{\sqrt x }} < 0\left( {do:\left\{ {5\sqrt x  + 2 > 0} \right.\forall x > 0} \right)\\
 \to 5\sqrt x  - 2 < 0\left( {do:\sqrt x  > 0\forall x > 0} \right)\\
 \to \sqrt x  < \dfrac{2}{5}\\
 \to 0 < x < \dfrac{4}{{25}}
\end{array}\)