Đáp án:
\(\dfrac{{12}}{{\sqrt a }}\)
Giải thích các bước giải:
\(\begin{array}{l}
Q = \left[ {\dfrac{{2\sqrt a - 3 + \sqrt a }}{{2\sqrt a \left( {3 - \sqrt a } \right)}}} \right].\left[ {\dfrac{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 3} \right) - \left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 1} \right)}}} \right].{\left( {\sqrt a - 3} \right)^2}\\
= - \dfrac{{3\sqrt a - 3}}{{2\sqrt a \left( {\sqrt a - 3} \right)}}.\dfrac{{a - 9 - a + 1}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 1} \right)}}.{\left( {\sqrt a - 3} \right)^2}\\
= - \dfrac{{3\left( {\sqrt a - 1} \right)}}{{2\sqrt a \left( {\sqrt a - 3} \right)}}.\dfrac{{ - 8}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 1} \right)}}.{\left( {\sqrt a - 3} \right)^2}\\
= \dfrac{{3\left( {\sqrt a - 1} \right)}}{{2\sqrt a \left( {\sqrt a - 3} \right)}}.\dfrac{8}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a - 1} \right)}}.{\left( {\sqrt a - 3} \right)^2}\\
= \dfrac{{12}}{{\sqrt a }}
\end{array}\)
