Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
B = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{8\sqrt x }}{{x - 1}}} \right):\left( {\dfrac{{\sqrt x - x - 3}}{{x - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right)\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right)\\
= \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{8\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\left( {\dfrac{{\sqrt x - x - 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{1}{{\sqrt x - 1}}} \right)\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2} - 8\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{\left( {\sqrt x - x - 3} \right) - \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {x + 2\sqrt x + 1} \right) - \left( {x - 2\sqrt x + 1} \right) - 8\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{\sqrt x - x - 3 - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{ - x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{ - x - 4}}\\
= \dfrac{{4\sqrt x }}{{x + 4}}\\
4,\\
A = \left( {\dfrac{1}{{1 - \sqrt x }} + \dfrac{1}{{1 + \sqrt x }}} \right):\left( {\dfrac{1}{{1 - \sqrt x }} - \dfrac{1}{{1 + \sqrt x }}} \right) + \dfrac{1}{{2\sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right)\\
= \dfrac{{\left( {1 + \sqrt x } \right) + \left( {1 - \sqrt x } \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}:\dfrac{{\left( {1 + \sqrt x } \right) - \left( {1 - \sqrt x } \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{1}{{2\sqrt x }}\\
= \dfrac{2}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}:\dfrac{{2\sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{1}{{2\sqrt x }}\\
= \dfrac{2}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{2\sqrt x }} + \dfrac{1}{{2\sqrt x }}\\
= \dfrac{1}{{\sqrt x }} + \dfrac{1}{{2\sqrt x }}\\
= \dfrac{3}{{2\sqrt x }}
\end{array}\)
