Đáp án:
d. \( - \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = 3 + 3\sqrt[3]{9}.\sqrt[3]{2} + 3.\sqrt[3]{3}.\sqrt[3]{4} + 2\\
= 5 + 3\sqrt[3]{9}.\sqrt[3]{2} + 3.\sqrt[3]{3}.\sqrt[3]{4}\\
= 5 + 3\sqrt[3]{3}\sqrt[3]{2}\left( {\sqrt[3]{3} + \sqrt[3]{2}} \right)\\
= 5 + 3\sqrt[3]{6}\left( {\sqrt[3]{3} + \sqrt[3]{2}} \right)\\
b.B = \left( {\sqrt[3]{5} - \sqrt[3]{3}} \right)\left( {\sqrt[3]{{{5^2}}} + \sqrt[3]{5}.\sqrt[3]{3} + \sqrt[3]{{{3^2}}}} \right)\\
= {\left( {\sqrt[3]{5}} \right)^3} - {\left( {\sqrt[3]{3}} \right)^3}\\
= 5 - 3 = 2\\
c.\sqrt[3]{{3.2.27}}.\sqrt[3]{{ - 2}}.\dfrac{{\sqrt[3]{2}}}{{\sqrt[3]{3}}}\\
= \sqrt[3]{{2.27}}.\sqrt[3]{{ - 2}}.\sqrt[3]{2}\\
= - \sqrt[3]{{{2^3}}}.\sqrt[3]{{{3^3}}}\\
= - 2.3 = - 6\\
d.\sqrt[3]{2}:\sqrt[3]{{{2^4}}} - \sqrt[3]{{\dfrac{{45}}{2}}}:\sqrt[3]{{\dfrac{{160}}{3}}}\\
= \dfrac{{\sqrt[3]{2}}}{{2.\sqrt[3]{2}}} - \dfrac{{\sqrt[3]{{3.3.5}}}}{{\sqrt[3]{2}}}.\dfrac{{\sqrt[3]{3}}}{{\sqrt[3]{{2.2.40}}}}\\
= \dfrac{1}{2} - \dfrac{{3\sqrt[3]{5}}}{{2\sqrt[3]{{40}}}} = \dfrac{1}{2} - \dfrac{{3\sqrt[3]{5}}}{{2\sqrt[3]{{{2^3}.5}}}}\\
= \dfrac{1}{2} - \dfrac{3}{{2.2}} = - \dfrac{1}{4}
\end{array}\)
