Giải Bài 3 trang 15 sách bài tập toán 7 tập 1 - Chân trời sáng tạo

Đề bài

Tính

a) ${\left( {\dfrac{{ - 2}}{3}} \right)^3}.{\left( {\dfrac{{ - 2}}{3}} \right)^2}$

b) ${(0,15)^7}:{(0,15)^5}$

c) ${\left( {\dfrac{3}{5}} \right)^{15}}:{\left( {\dfrac{{27}}{{125}}} \right)^5}$

d) ${\left( {\dfrac{1}{7}} \right)^4}.\dfrac{1}{7}{.49^3}$

Lời giải chi tiết

$\begin{array}{l}a){\left( {\dfrac{{ - 2}}{3}} \right)^3}.{\left( {\dfrac{{ - 2}}{3}} \right)^2} = {\left( {\dfrac{{ - 2}}{3}} \right)^{3 + 2}} = {\left( {\dfrac{{ - 2}}{3}} \right)^5}\\b){(0,15)^7}:{(0,15)^5} = {(0,15)^{7 - 5}} = {(0,15)^2}\\c){\left( {\dfrac{3}{5}} \right)^{15}}:{\left( {\dfrac{{27}}{{125}}} \right)^5} = {\left( {\dfrac{3}{5}} \right)^{15}}:{\left( {\dfrac{{{3^3}}}{{{5^3}}}} \right)^5} = {\left( {\dfrac{3}{5}} \right)^{15}}:{\left( {\dfrac{3}{5}} \right)^{5.3}} = {\left( {\dfrac{3}{5}} \right)^{15 - 15}} = {\left( {\dfrac{3}{5}} \right)^0} = 1\\d){\left( {\dfrac{1}{7}} \right)^4}.\dfrac{1}{7}{.49^3} = \dfrac{1}{{{7^4}}}.\dfrac{1}{7}.{(7^2)^3} = \dfrac{1}{{{7^4}}}.\dfrac{1}{7}{.7^6} = \dfrac{{{{1.1.7}^6}}}{{{7^4}.7}} = 7\end{array}$