Đáp án:
$\begin{array}{l}
1)a)A = \sqrt[3]{{{{\left( {1 - \sqrt 6 } \right)}^3}}} + \dfrac{{\sqrt[3]{{320}}}}{{\sqrt[3]{5}}} - \sqrt[3]{{27}}\\
= 1 - \sqrt 6 + \sqrt[3]{{\dfrac{{320}}{5}}} - \sqrt[3]{{{3^3}}}\\
= 1 - \sqrt 6 + \sqrt[3]{{64}} - 3\\
= - 2 - \sqrt 6 + \sqrt[3]{{{4^3}}}\\
= - 2 - \sqrt 6 + 4\\
= 2 - \sqrt 6 \\
b)\\
B = \sqrt {24 - 16\sqrt 2 } + \dfrac{4}{{\sqrt 2 - 1}} + \sqrt {98} \\
= \sqrt {16 - 2.4.2\sqrt 2 + 8} + \dfrac{{4\left( {\sqrt 2 + 1} \right)}}{{2 - 1}} + \sqrt {49.2} \\
= \sqrt {{{\left( {4 - 2\sqrt 2 } \right)}^2}} + 4\left( {\sqrt 2 + 1} \right) + 7\sqrt 2 \\
= 4 - 2\sqrt 2 + 4\sqrt 2 + 4 + 7\sqrt 2 \\
= 8 + 9\sqrt 2 \\
2)a)\sqrt[3]{{2x - 1}} - \sqrt[3]{{16x - 8}} = 3\\
\Leftrightarrow \sqrt[3]{{2x - 1}} - \sqrt[3]{{8\left( {2x - 1} \right)}} = 3\\
\Leftrightarrow \sqrt[3]{{2x - 1}} - 2\sqrt[3]{{2x - 1}} = 3\\
\Leftrightarrow - \sqrt[3]{{2x - 1}} = 3\\
\Leftrightarrow \sqrt[3]{{2x - 1}} = - 3\\
\Leftrightarrow 2x - 1 = {\left( { - 3} \right)^3}\\
\Leftrightarrow 2x - 1 = - 27\\
\Leftrightarrow 2x = - 26\\
\Leftrightarrow x = - 13\\
Vậy\,x = - 13\\
b)Dkxd:x \ge - 3\\
3.\sqrt {\dfrac{{x + 3}}{9}} - \sqrt {16x + 48} + 3 = 0\\
\Leftrightarrow 3.\dfrac{1}{3}\sqrt {x + 3} - 4\sqrt {x + 3} = - 3\\
\Leftrightarrow \sqrt {x + 3} - 4\sqrt {x + 3} = - 3\\
\Leftrightarrow - 3\sqrt {x + 3} = - 3\\
\Leftrightarrow \sqrt {x + 3} = 1\\
\Leftrightarrow x + 3 = 1\\
\Leftrightarrow x = - 2\left( {tmdk} \right)\\
Vậy\,x = - 2
\end{array}$
