Viết mỗi biểu thức sau dưới dạng một lũy thừa \(\left( {a > 0} \right)\):
a) \(\sqrt[4]{{{2^{ - 3}}}}\);
b) \(\frac{1}{{\sqrt[5]{{{2^3}}}}}\);
c) \({\left( {\sqrt[5]{3}} \right)^4}\);
d) \(\sqrt {a\sqrt[3]{a}} \);
e) \(\sqrt[3]{a}.\sqrt[4]{{{a^3}}}:{\left( {\sqrt[6]{a}} \right)^5}\);
g) \({a^{\frac{1}{3}}}:{a^{ - \frac{3}{2}}}.{a^{ - \frac{2}{3}}}\).
Sử dụng kiến thức về phép tính lũy thừa để tính:
a, c, d) \({\left( {\sqrt[n]{a}} \right)^m} = \sqrt[n]{{{a^m}}}\), \({\left( {{a^\alpha }} \right)^\beta } = {a^{\alpha \beta }}\), \({a^{\frac{m}{n}}} = \sqrt[n]{{{a^m}}}\)
b) Với \(m,n \in \mathbb{Z},n > 0\) thì: \({a^{\frac{m}{n}}} = \sqrt[n]{{{a^m}}}\), \({a^{ - n}} = \frac{1}{{{a^n}}}\) \(\left( {a \ne 0} \right)\).
e, g) \({\left( {\sqrt[n]{a}} \right)^m} = \sqrt[n]{{{a^m}}}\), \({a^{\frac{m}{n}}} = \sqrt[n]{{{a^m}}}\), \({a^\alpha }.{a^\beta } = {a^{\alpha + \beta }}\), \({\left( {\frac{a}{b}} \right)^\alpha } = \frac{{{a^\alpha }}}{{{b^\alpha }}}\)
a) \(\sqrt[4]{{{2^{ - 3}}}} = {\left( {\sqrt[4]{2}} \right)^{ - 3}} = {2^{\frac{{ - 3}}{4}}}\);
b) \(\frac{1}{{\sqrt[5]{{{2^3}}}}} = \frac{1}{{{2^{\frac{3}{5}}}}} = {2^{ - \frac{3}{5}}}\);
c) \({\left( {\sqrt[5]{3}} \right)^4} = {\left( {{3^{\frac{1}{5}}}} \right)^4} = {3^{\frac{4}{5}}}\);
d) \(\sqrt {a\sqrt[3]{a}} = {\left( {a.{a^{\frac{1}{3}}}} \right)^{\frac{1}{2}}} = {\left( {{a^{\frac{4}{3}}}} \right)^{\frac{1}{2}}} = {a^{\frac{2}{3}}}\);
e) \(\sqrt[3]{a}.\sqrt[4]{{{a^3}}}:{\left( {\sqrt[6]{a}} \right)^5} = {a^{\frac{1}{3}}}.{a^{\frac{3}{4}}}:{a^{\frac{5}{6}}} = {a^{\frac{1}{3} + \frac{3}{4} - \frac{5}{6}}} = {a^{\frac{1}{4}}}\);
g) \({a^{\frac{1}{3}}}:{a^{ - \frac{3}{2}}}.{a^{ - \frac{2}{3}}} = {a^{\frac{1}{3} - \left( { - \frac{3}{2}} \right) + \left( {\frac{{ - 2}}{3}} \right)}} = {a^{\frac{7}{6}}}\).