Đáp án:
$\begin{array}{l}
2)A = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{x - 2\sqrt x - 3}}{{x - 1}}} \right):\left( {\dfrac{{x + 3}}{{x - 1}} + \dfrac{2}{{\sqrt x + 1}}} \right)\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - x + 2\sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}:\dfrac{{x + 3 + 2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x + 3}}{{\left( {\sqrt x + 1} \right).\left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x + 3 + 2\sqrt x - 2}}\\
= \dfrac{{4\sqrt x + 4}}{1}.\dfrac{1}{{x + 2\sqrt x + 1}}\\
= \dfrac{{4\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\
= \dfrac{4}{{\sqrt x + 1}}\\
2)a)m = 1\\
\Leftrightarrow {x^2} - 2x - 1 = 0\\
\Leftrightarrow {x^2} - 2x + 1 = 2\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 2\\
\Leftrightarrow x = 1 \pm \sqrt 2 \\
Vậy\,x = 1 \pm \sqrt 2 \\
b)\Delta ' > 0\\
\Leftrightarrow {m^2} - {m^2} + 2 > 0\\
\Leftrightarrow 2 > 0\left( {tmdk} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m\\
{x_1}{x_2} = {m^2} - 2
\end{array} \right.\\
\left| {x_1^3 - x_2^3} \right| = 10\sqrt 2 \\
\Leftrightarrow \left| {\left( {{x_1} - {x_2}} \right)\left( {x_1^2 + {x_1}{x_2} + x_2^2} \right)} \right| = 10\sqrt 2 \\
\Leftrightarrow {\left( {{x_1} - {x_2}} \right)^2}{\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - {x_1}{x_2}} \right]^2} = 200\\
\Leftrightarrow \left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \right].{\left( {4{m^2} - {m^2} + 2} \right)^2} = 200\\
\Leftrightarrow \left( {4{m^2} - 4{m^2} + 8} \right){\left( {3{m^2} + 2} \right)^2} = 200\\
\Leftrightarrow {\left( {3{m^2} + 2} \right)^2} = 15\\
\Leftrightarrow 3{m^2} + 2 = \sqrt {15} \\
\Leftrightarrow {m^2} = \dfrac{{\sqrt {15} - 2}}{3}\\
\Leftrightarrow m = \pm \dfrac{{\sqrt {\sqrt {45} - 6} }}{3}\\
Vậy\,m = \pm \dfrac{{\sqrt {\sqrt {45} - 6} }}{3}
\end{array}$
