Tính
a) \({\left( {\dfrac{{ - 1}}{3}} \right)^4}\),\({\left( {\dfrac{{ - 2}}{3}} \right)^3}\),\({\left( {2\dfrac{1}{2}} \right)^3}\),\({\left( { - 0,2} \right)^3}\)
b) \({\left( {\dfrac{{ - 1}}{2}} \right)^2}\),\({\left( {\dfrac{{ - 1}}{2}} \right)^3}\),\({\left( {\dfrac{{ - 1}}{2}} \right)^4}\),\({\left( {\dfrac{{ - 1}}{2}} \right)^5}\)
Ta sử dụng định nghĩa lũy thừa của 1 số hữu tỉ
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\(\begin{array}{l}a){\left( {\dfrac{{ - 1}}{3}} \right)^4} = \dfrac{{{{( - 1)}^4}}}{{{3^4}}} = \dfrac{{( - 1).( - 1).( - 1).( - 1)}}{{3.3.3.3}} = \dfrac{1}{{81}}\\{\left( {\dfrac{{ - 2}}{3}} \right)^3} = \dfrac{{{{( - 2)}^3}}}{{{3^3}}} = \dfrac{{( - 2).( - 2).( - 2)}}{{3.3.3}} = \dfrac{{ - 8}}{{27}}\end{array}\)
\({\left( {2\dfrac{1}{2}} \right)^3} = {\left( {\dfrac{5}{2}} \right)^3} = \dfrac{{5.5.5}}{{2.2.2}} = \dfrac{{125}}{8}\)
\({\left( { - 0,2} \right)^3} = {\left( {\dfrac{{ - 1}}{5}} \right)^3} = \dfrac{{( - 1).( - 1).( - 1)}}{{5.5.5}} = \dfrac{{ - 1}}{{125}}\)
\(\begin{array}{l}b){\left( {\dfrac{{ - 1}}{2}} \right)^2} = \dfrac{{( - 1).( - 1)}}{{2.2}} = \dfrac{1}{4}\\{\left( {\dfrac{{ - 1}}{2}} \right)^3} = \dfrac{{( - 1).( - 1).( - 1)}}{{2.2.2}} = \dfrac{{ - 1}}{8}\\{\left( {\dfrac{{ - 1}}{2}} \right)^4} = \dfrac{{( - 1).( - 1).( - 1).( - 1)}}{{2.2.2.2}} = \dfrac{1}{{16}}\\{\left( {\dfrac{{ - 1}}{2}} \right)^5} = \dfrac{{( - 1).( - 1).( - 1).( - 1).( - 1)}}{{2.2.2.2.2}} = \dfrac{{ - 1}}{{32}}\end{array}\)