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