Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = 3\sqrt {12} - 4\sqrt {27} + 5\sqrt {48} \\
= 3\sqrt {{2^2}.3} - 4.\sqrt {{3^2}.3} + 5.\sqrt {{4^2}.3} \\
= 3.2.\sqrt 3 - 4.3.\sqrt 3 + 5.4.\sqrt 3 \\
= 6\sqrt 3 - 12\sqrt 3 + 20\sqrt 3 \\
= 14\sqrt 3 \\
b,\\
B = \dfrac{{\sqrt a }}{{\sqrt a - 3}} - \dfrac{3}{{\sqrt a + 3}} - \dfrac{{a - 2}}{{a - 9}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
a \ge 0\\
a \ne 9
\end{array} \right)\\
= \dfrac{{\sqrt a }}{{\sqrt a - 3}} - \dfrac{3}{{\sqrt a + 3}} - \dfrac{{a - 2}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)}}\\
= \dfrac{{\sqrt a .\left( {\sqrt a + 3} \right) - 3.\left( {\sqrt a - 3} \right) - \left( {a - 2} \right)}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)}}\\
= \dfrac{{a + 3\sqrt a - 3\sqrt a + 9 - a - 2}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)}}\\
= \dfrac{7}{{a - 9}}
\end{array}\)
