Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# \dfrac{1}{2}\\
A = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt {2x} + 1}} + \dfrac{{\sqrt {2x} + \sqrt x }}{{\sqrt {2x} - 1}} - 1} \right)\\
:\left( {1 + \dfrac{{\sqrt x + 1}}{{\sqrt {2x} + 1}} - \dfrac{{\sqrt {2x} + \sqrt x }}{{\sqrt {2x} - 1}}} \right)\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt {2x} - 1} \right) + \left( {\sqrt {2x} + \sqrt x } \right)\left( {\sqrt {2x} + 1} \right) - \left( {\sqrt {2x} + 1} \right)\left( {\sqrt {2x} - 1} \right)}}{{\left( {\sqrt {2x} + 1} \right)\left( {\sqrt {2x} - 1} \right)}}\\
:\dfrac{{\left( {\sqrt {2x} + 1} \right)\left( {\sqrt {2x} - 1} \right) + \left( {\sqrt x + 1} \right)\left( {\sqrt {2x} - 1} \right) - \left( {\sqrt {2x} + \sqrt x } \right)\left( {\sqrt {2x} + 1} \right)}}{{\left( {\sqrt {2x} + 1} \right)\left( {\sqrt {2x} - 1} \right)}}\\
= \dfrac{{\sqrt 2 x + \left( {\sqrt 2 - 1} \right).\sqrt x - 1 + \left( {2 + \sqrt 2 } \right).x + \left( {\sqrt 2 + 1} \right).\sqrt x - 2x + 1}}{{2x - 1}}\\
.\dfrac{{2x - 1}}{{2x - 1 + \sqrt 2 x + \left( {\sqrt 2 - 1} \right).\sqrt x - 1 - \left( {2 + \sqrt 2 } \right).x - \left( {\sqrt 2 + 1} \right).\sqrt x }}\\
= \dfrac{{2\sqrt 2 .x + 2\sqrt 2 .\sqrt x }}{{ - 2\sqrt x - 2}}\\
= \dfrac{{2\sqrt 2 .\sqrt x \left( {\sqrt x + 1} \right)}}{{ - 2\left( {\sqrt x + 1} \right)}}\\
= - \sqrt {2x} \\
b)x = \dfrac{{3 + 2\sqrt 2 }}{2}\left( {tmdk} \right)\\
= \dfrac{{{{\left( {\sqrt 2 + 1} \right)}^2}}}{2}\\
\Leftrightarrow \sqrt x = \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}\\
\Leftrightarrow A = - \sqrt {2x} \\
= - \sqrt {2.\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}} \\
= - \sqrt {2 + \sqrt 2 }
\end{array}$
