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

Đề bài

Tính giá trị các biểu thức

a) \(\dfrac{{{9^3}{{.2}^{10}}}}{{{{16}^2}{{.81}^2}}}\)

b) \(\dfrac{{{{( - 3)}^7}.{{( - 3)}^8}}}{{{{7.9}^7}}}\)

c) \(\dfrac{{{{(0,3)}^6}.{{(0,04)}^3}}}{{{{(0,09)}^4}.{{(0,2)}^4}}}\)

d) \(\dfrac{{{2^3} + {2^4} + {2^5} + {2^6}}}{{{{15}^2}}}\)

Lời giải chi tiết

\(\begin{array}{l}a)\dfrac{{{9^3}{{.2}^{10}}}}{{{{16}^2}{{.81}^2}}} = \dfrac{{{9^3}{{.2}^{10}}}}{{{{({2^4})}^2}.{{({9^2})}^2}}} = \dfrac{{{9^3}{{.2}^{10}}}}{{{2^8}{{.9}^4}}} = \dfrac{{{2^2}}}{9} = \dfrac{4}{9}\\b)\dfrac{{{{( - 3)}^7}.{{( - 3)}^8}}}{{{{7.9}^7}}} = \dfrac{{{{( - 3)}^{15}}}}{{7.{{\left[ {{{\left( { - 3} \right)}^2}} \right]}^7}}} = \dfrac{{{{( - 3)}^{15}}}}{{7.{{( - 3)}^{14}}}} = \dfrac{{ - 3}}{7}\\c)\dfrac{{{{(0,3)}^6}.{{(0,04)}^3}}}{{{{(0,09)}^4}.{{(0,2)}^4}}} = \dfrac{{{{(0,3)}^6}.{{(0,{2^2})}^3}}}{{{{(0,{3^2})}^4}.{{(0,2)}^4}}} = \dfrac{{{{(0,3)}^6}.{{(0,2)}^6}}}{{{{(0,3)}^8}.{{(0,2)}^4}}} = \dfrac{{{{(0,2)}^2}}}{{{{(0,3)}^2}}} = \dfrac{{0,04}}{{0,09}} = \dfrac{4}{9}\\d)\dfrac{{{2^3} + {2^4} + {2^5} + {2^6}}}{{{{15}^2}}} = \dfrac{{8 + 16 + 32 + 64}}{{225}} = \dfrac{{120}}{{225}} = \dfrac{8}{{15}}\end{array}\)