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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