Điền dấu >, < hoặc = và chỗ chấm (…) :
a) \(\sqrt[3]{{64}}...\sqrt {64} \) b) \(\sqrt[3]{{0,001}}...0,01\);
c) \( - \dfrac{1}{4}...\sqrt[3]{{ - \dfrac{8}{{125}}}}\).
Sử dụng công thức: \(a < b \Leftrightarrow \sqrt[3]{a} < \sqrt[3]{b}.\)
\(a)\;\sqrt[3]{{64}}...\sqrt {64} \)
Ta có: \(\sqrt[3]{{64}} = \sqrt[3]{{{4^3}}} = 4;\;\sqrt {64} = \sqrt {{8^2}} = 8.\)
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Vì \(4 < 8 \Rightarrow \sqrt[3]{{64}} < \sqrt {64} .\)
Vậy \(\sqrt[3]{{64}} < \sqrt {64} .\)
\(b)\;\sqrt[3]{{0,001}}...0,01\)
Ta có : \(\sqrt[3]{{0,001}} = \sqrt[3]{{0,{1^3}}} = 0,1.\)
Vì \(0,1 > 0,01 \Rightarrow \sqrt[3]{{0,001}} > 0,01.\)
Vậy \(\sqrt[3]{{0,001}} > 0,01.\)
\(c)\; - \dfrac{1}{4}...\sqrt[3]{{ - \dfrac{8}{{125}}}}\)
Ta có: \(\sqrt[3]{{ - \dfrac{8}{{125}}}} = \sqrt[3]{{{{\left( { - \dfrac{2}{5}} \right)}^3}}} = - \dfrac{2}{5} = - \dfrac{8}{{20}};\;\; - \dfrac{1}{4} = - \dfrac{5}{{20}}.\)
Vì \( - \dfrac{5}{{20}} > - \dfrac{8}{{20}} \Rightarrow - \dfrac{1}{4} > \sqrt[3]{{ - \dfrac{8}{{125}}}}.\)
Vậy \( - \dfrac{1}{4} > \sqrt[3]{{ - \dfrac{8}{{125}}}}.\)