Bài 27 trang 19 SGK Toán 7 tập 1

Đề bài

Tính: \({\left( {\dfrac{{ - 1}}{3}} \right)^4};\,\,{\left( { - 2\dfrac{1}{4}} \right)^3};\,\,{\left( { - 0,2} \right)^2};\,{\left( { - 5,3} \right)^0}\)

Lời giải chi tiết

+) \({\left( {\dfrac{{ - 1}}{3}} \right)^4} = \left( {\dfrac{{ - 1}}{3}} \right).\left( {\dfrac{{ - 1}}{3}} \right).\left( {\dfrac{{ - 1}}{3}} \right).\left( {\dfrac{{ - 1}}{3}} \right) \)\(= \dfrac{1}{{81}}\)

+) \({\left( { - 2\dfrac{1}{4}} \right)^3} \)\(= {\left( {\dfrac{{ - 9}}{4}} \right)^3} \)\(= \left( {\dfrac{{ - 9}}{4}} \right).\left( {\dfrac{{ - 9}}{4}} \right).\left( {\dfrac{{ - 9}}{4}} \right) \)\(= \dfrac{{ - 729}}{{64}}\)

+) \(\,{\left( { - 0,2} \right)^2} = \left( { - 0,2} \right).\left( { - 0,2} \right) = 0,04\)

+) \({\left( { - 5,3} \right)^0} = 1\)

Cách khác cho 2 ý đầu:

\(\begin{array}{l}
{\left( {\dfrac{{ - 1}}{3}} \right)^4} = \dfrac{{{{\left( { - 1} \right)}^4}}}{{{3^4}}} = \dfrac{1}{{81}}\\
{\left( { - 2\dfrac{1}{4}} \right)^3} = {\left( {\dfrac{{ - 9}}{4}} \right)^3} = \dfrac{{{{\left( { - 9} \right)}^3}}}{{{4^3}}} = \dfrac{{ - 729}}{{64}}
\end{array}\)

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