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: \(\sqrt[3]{{64}} = \sqrt[3]{{{4^3}}} = 4;\;\sqrt {64} = \sqrt {{8^2}} = 8.\)
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 \( – \dfrac{1}{4} > \sqrt[3]{{ – \dfrac{8}{{125}}}}.\)