Giải bài 3 trang 27 SGK Toán 7 tập 1 - Chân trời sáng tạo

Đề bài

Thực hiện phép tính.

a) \(\frac{{{5^{16}}{{.27}^7}}}{{{{125}^5}{{.9}^{11}}}}\)         b)\({\left( { - 0,2} \right)^2}.5 - \frac{{{2^{13}}{{.27}^3}}}{{{4^6}{{.9}^5}}};\)

c)\(\frac{{{5^6} + {2^2}{{.25}^3} + {2^3}{{.125}^2}}}{{{{26.5}^6}}}.\)

Lời giải chi tiết

a)

\(\frac{{{5^{16}}{{.27}^7}}}{{{{125}^5}{{.9}^{11}}}} = \frac{{{5^{16}}.{{\left( {{3^3}} \right)}^7}}}{{{{\left( {{5^3}} \right)}^5}.{{\left( {{3^2}} \right)}^{11}}}} = \frac{{{5^{16}}{{.3}^{21}}}}{{{5^{15}}{{.3}^{22}}}} = \frac{5.{{5^{15}}{{.3}^{21}}}}{{{5^{15}}{.3{.3}^{21}}}}= \frac{5}{3}\)

b)

\({\left( { - 0,2} \right)^2}.5 - \frac{{{2^{13}}{{.27}^3}}}{{{4^6}{{.9}^5}}} = 0,04.5 - \frac{{{2^{13}}.{{\left( {{3^3}} \right)}^3}}}{{{{\left( {{2^2}} \right)}^6}.{{\left( {{3^2}} \right)}^5}}}\\ = 0,2 - \frac{{{2^{13}}{{.3}^9}}}{{{2^{12}}{{.3}^{10}}}}\\ = \frac{1}{5} - \frac{2}{{3}} \\= \frac{3}{15} - \frac{10}{{15}}\\ = \frac{{-7}}{{15}}\)

c)

\(\begin{array}{l}\frac{{{5^6} + {2^2}{{.25}^3} + {2^3}{{.125}^2}}}{{{{26.5}^6}}} = \frac{{{5^6} + {2^2}.{{\left( {{5^2}} \right)}^3} + {2^3}.{{\left( {{5^3}} \right)}^2}}}{{{{2.13.5}^6}}}\\ = \frac{{{5^6} + {{4.5}^6} + {{8.5}^6}}}{{{{2.13.5}^6}}} = \frac{{{5^6}.\left( {1 + 4 + 8} \right)}}{{{{2.13.5}^6}}}\\ = \frac{{{5^6}.13}}{{{{2.13.5}^6}}} = \frac{1}{2}\end{array}\)