Giải bài 8 trang 21 SGK Toán 7 tập 1 - Chân trời sáng tạo — Không quảng cáo

Đề bài

Tính giá trị các biểu thức.

a)\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}};\)

b)\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}};\)

c)\(\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}};\)

d)\(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}}.\)

Lời giải chi tiết

a) \(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}} \) \(= \frac{{{{\left( {{2^2}} \right)}^3}.{{\left( {{3^2}} \right)}^7}}}{{{{\left( {{3^3}} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} \) \(=\frac{2^{2.3}.3^{2.7}}{3^{3.5}.2^{2.3}}\) \(= \frac{{{2^6}{{.3}^{14}}}}{{{3^{15}}{{.2}^6}}} \) \(= \frac{1}{3}\)

b) \(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}} \) \(=\frac{(-2)^{3+7}}{3.(2^2)^6}\) \(= \frac{{{{\left( { - 2} \right)}^{10}}}}{{3.{{\left( {{2^{2.6}}} \right)}}}} \) \(= \frac{{{2^{10}}}}{{{{3.2}^{12}}}} \) \(= \frac{1}{{{{3.2}^2}}} \) \(= \frac{1}{{12}}\)

c) \(\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} \) \(= \frac{{{{\left( {0,2} \right)}^5}.{{\left[ {{{\left( {0,3} \right)}^2}} \right]}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} \) \(= \frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,3} \right)}^6}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}}\\ \) \(= \frac{{{{\left( {0,3} \right)}^2}}}{{{{\left( {0,2} \right)}^2}}} \) \(= \frac{{0,09}}{{0,04}} \) \(= \frac{9}{4}\)

d)

Cách 1: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} \) \(= \frac{{8 + 16 + 32}}{{49}} \) \(= \frac{{56}}{{49}} \) \(= \frac{8}{7}\)

Cách 2: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} \) \(= \frac{{2^3.(1+2+2^2)}}{{7^2}} \) \(= \frac{{2^3.7}}{{7^2}} \) \(= \frac{8}{7}\)