Câu 137212:
A. \(a)\;\frac{{\sqrt[3]{4}}}{4},\;\;b)\;1 + \sqrt[3]{2} + \sqrt[3]{4},\;c)\;\frac{{\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}}}{5}\)
B. \(a)\;\frac{{\sqrt[3]{4}}}{4},\;\;b)\;1 + \sqrt[3]{2} + \sqrt[3]{4},\;c)\;\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}\)
C. \(a)\;\frac{{\sqrt[3]{4}}}{4},\;\;b)\; - 1 - \sqrt[3]{2} - \sqrt[3]{4},\;c)\;\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}\)
D. \(a)\;\frac{{\sqrt[3]{4}}}{4},\;\;b)\; - 1 - \sqrt[3]{2} - \sqrt[3]{4},\;c)\;\frac{{\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}}}{5}\)
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Đáp án : D(14) bình luận (0) lời giải
Giải chi tiết:
\(\begin{array}{l}
a)\;\;\frac{1}{{2\sqrt[3]{2}}} = \frac{{{{\left( {\sqrt[3]{2}} \right)}^2}}}{{2{{\left( {\sqrt[3]{2}} \right)}^3}}} = \frac{{\sqrt[3]{4}}}{4}.\\
b)\;\;\frac{1}{{1 - \sqrt[3]{2}}} = \frac{{1 + \sqrt[3]{2} + \sqrt[3]{4}}}{{\left( {1 - \sqrt[3]{2}} \right)\left( {1 + \sqrt[3]{2} + \sqrt[3]{4}} \right)}}\\
\;\;\; = \frac{{1 + \sqrt[3]{2} + \sqrt[3]{4}}}{{1 - 2}} = - \left( {1 + \sqrt[3]{2} + \sqrt[3]{4}} \right).\\
c)\;\frac{1}{{\sqrt[3]{3} + \sqrt[3]{2}}} = \frac{{\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}}}{{\left( {\sqrt[3]{3} + \sqrt[3]{2}} \right)\left( {\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}} \right)}}\\
\; = \frac{{\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}}}{{3 + 2}} = \frac{{\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}}}{5}.
\end{array}\)Lời giải sai Bình thường Khá hay Rất Hay
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