Đáp án:
$\begin{array}{l}
c)Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}}\\
A = \dfrac{1}{4}\\
\Leftrightarrow \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{1}{4}\\
\Leftrightarrow 8\sqrt x - 4 = \sqrt x + 1\\
\Leftrightarrow 7\sqrt x = 5\\
\Leftrightarrow \sqrt x = \dfrac{5}{7}\\
\Leftrightarrow x = \dfrac{{25}}{{49}}\left( {tmdk} \right)\\
Vay\,x = \dfrac{{25}}{{49}}\\
e)m.A = \sqrt x - 2\\
\Leftrightarrow m.\dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}} = \sqrt x - 2\\
\Leftrightarrow 2m\sqrt x - m = \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)\\
\Leftrightarrow 2m\sqrt x - m = x - \sqrt x - 2\\
\Leftrightarrow x - \left( {2m + 1} \right).\sqrt x + m - 2 = 0\\
Dat:\sqrt x = t\left( {t \ge 0;t \ne 1} \right)\\
\Leftrightarrow {t^2} - \left( {2m + 1} \right).t + m - 2 = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\Delta > 0\\
{t_1} + {t_2} > 0\\
{t_1}.{t_2} > 0\\
{1^2} - \left( {2m + 1} \right).1 + m - 2 \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {2m + 1} \right)^2} - 4\left( {m - 2} \right) > 0\\
2m + 1 > 0\\
m - 2 > 0\\
1 - 2m - 1 + m - 2 \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4{m^2} + 4m + 1 - 4m + 8 > 0\\
m > 2\\
m \ne - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4{m^2} + 9 > 0\left( {tm} \right)\\
m > 2
\end{array} \right.\\
\Leftrightarrow m > 2\\
Vay\,m > 2\\
f)A < 1\\
\Leftrightarrow \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}} - 1 < 0\\
\Leftrightarrow \dfrac{{2\sqrt x - 1 - \sqrt x - 1}}{{\sqrt x + 1}} < 0\\
\Leftrightarrow \sqrt x - 2 < 0\\
\Leftrightarrow \sqrt x < 2\\
\Leftrightarrow x < 4\\
Vay\,0 \le x < 4;x \ne 1
\end{array}$
