Bài 2 trang 14 SBT Toán lớp 7 tập 1 Chân trời sáng tạo: Tính...

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

$\begin{array}{l}a){\left( {\dfrac{{ - 1}}{3}} \right)^4} = \dfrac{{{{( - 1)}^4}}}{{{3^4}}} = \dfrac{{( - 1).( - 1).( - 1).( - 1)}}{{3.3.3.3}} = \dfrac{1}{{81}}\\{\left( {\dfrac{{ - 2}}{3}} \right)^3} = \dfrac{{{{( - 2)}^3}}}{{{3^3}}} = \dfrac{{( - 2).( - 2).( - 2)}}{{3.3.3}} = \dfrac{{ - 8}}{{27}}\end{array}$

${\left( {2\dfrac{1}{2}} \right)^3} = {\left( {\dfrac{5}{2}} \right)^3} = \dfrac{{5.5.5}}{{2.2.2}} = \dfrac{{125}}{8}$

${\left( { - 0,2} \right)^3} = {\left( {\dfrac{{ - 1}}{5}} \right)^3} = \dfrac{{( - 1).( - 1).( - 1)}}{{5.5.5}} = \dfrac{{ - 1}}{{125}}$

$\begin{array}{l}b){\left( {\dfrac{{ - 1}}{2}} \right)^2} = \dfrac{{( - 1).( - 1)}}{{2.2}} = \dfrac{1}{4}\\{\left( {\dfrac{{ - 1}}{2}} \right)^3} = \dfrac{{( - 1).( - 1).( - 1)}}{{2.2.2}} = \dfrac{{ - 1}}{8}\\{\left( {\dfrac{{ - 1}}{2}} \right)^4} = \dfrac{{( - 1).( - 1).( - 1).( - 1)}}{{2.2.2.2}} = \dfrac{1}{{16}}\\{\left( {\dfrac{{ - 1}}{2}} \right)^5} = \dfrac{{( - 1).( - 1).( - 1).( - 1).( - 1)}}{{2.2.2.2.2}} = \dfrac{{ - 1}}{{32}}\end{array}$