Đáp án:
$\begin{array}{l}
1)a)A = \left( {3\sqrt {18} + 2\sqrt {50} - 4\sqrt {72} } \right):8\sqrt 2 \\
= \left( {3.3\sqrt 2 + 2.5\sqrt 2 - 4.6\sqrt 2 } \right):8\sqrt 2 \\
= \left( {9\sqrt 2 } \right):8\sqrt 2 \\
= \dfrac{9}{8}\\
b)B = \left( { - 4\sqrt {20} + 5\sqrt {500} - 3\sqrt {45} } \right):\sqrt 5 \\
= \left( { - 4.2\sqrt 5 + 5.10\sqrt 5 - 3.3\sqrt 5 } \right):\sqrt 5 \\
= \left( { - 8\sqrt 5 + 50\sqrt 5 - 9\sqrt 5 } \right):\sqrt 5 \\
= 33\sqrt 5 :\sqrt 5 \\
= 33\\
c)C = \left( {\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} - \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right):\sqrt {48} \\
= \dfrac{{{{\left( {\sqrt 3 + 1} \right)}^2} - {{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}:4\sqrt 3 \\
= \dfrac{{4 + 2\sqrt 3 - 4 + 2\sqrt 3 }}{{3 - 1}}:4\sqrt 3 \\
= 2\sqrt 3 :4\sqrt 3 \\
= \dfrac{1}{2}\\
B2)\\
a)Dkxd:x \ge 0;x \ne 9\\
b)A = 3x - \sqrt {48} + \dfrac{{\sqrt {{x^3}} - 3x}}{{\sqrt x - 3}}\\
= 3x - \sqrt {48} + \dfrac{{x\sqrt x - 3x}}{{\sqrt x - 3}}\\
= 3x - \sqrt {48} + \dfrac{{x\left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}}\\
= 3x - 4\sqrt 3 + x\\
= 4x - 4\sqrt 3
\end{array}$
