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