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