Đáp án:
$\begin{array}{l}
A = \frac{{\sqrt {x - 1} + 1}}{{2\sqrt {x - 1} + 3}}\\
a)Đkxđ:\left\{ \begin{array}{l}
x - 1 \ge 0\\
2\sqrt {x - 1} + 3 \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x \ge 1\\
\sqrt {x - 1} \ne - \frac{3}{2}\left( {tm} \right)
\end{array} \right.\\
\Rightarrow x \ge 1\\
b)A = \frac{2}{5}\\
\Rightarrow \frac{{\sqrt {x - 1} + 1}}{{2\sqrt {x - 1} + 3}} = \frac{2}{5}\\
\Rightarrow 5\left( {\sqrt {x - 1} + 1} \right) = 2\left( {2\sqrt {x - 1} + 3} \right)\\
\Rightarrow 5\sqrt {x - 1} + 5 = 4\sqrt {x - 1} + 6\\
\Rightarrow \sqrt {x - 1} = 1\\
\Rightarrow x - 1 = 1\\
\Rightarrow x = 2\left( {tmdk:x \ge 1} \right)\\
Vậy\,x = 2\,thì\,A = \frac{2}{5}
\end{array}$
