Giải bài 7 trang 21 SGK Toán 7 tập 1 – Chân trời sáng tạo

Đề bài

Tính:

a)\({\left( {\frac{2}{5} + \frac{1}{2}} \right)^2}\);                 b)\({\left( {0,75 – 1\frac{1}{2}} \right)^3};\)

c)\({\left( {\frac{3}{5}} \right)^{15}}:{\left( {0,36} \right)^5}\);         d)\({\left( {1 – \frac{1}{3}} \right)^8}:{\left( {\frac{4}{9}} \right)^3}\)

Lời giải chi tiết

a)\({\left( {\frac{2}{5} + \frac{1}{2}} \right)^2} = {\left( {\frac{4}{{10}} + \frac{5}{{10}}} \right)^2} = {\left( {\frac{9}{{10}}} \right)^2} = \frac{{81}}{{100}}\);               

 b)\({\left( {0,75 – 1\frac{1}{2}} \right)^3} = {\left( {\frac{3}{4} – \frac{3}{2}} \right)^3} = {\left( {\frac{3}{4} – \frac{6}{4}} \right)^3} = {\left( { – \frac{3}{4}} \right)^3} = \frac{{ – 27}}{{64}};\)

c)

\(\begin{array}{l}{\left( {\frac{3}{5}} \right)^{15}}:{\left( {0,36} \right)^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{9}{{25}}} \right)^5}\\ = {\left( {\frac{3}{5}} \right)^{15}}:{\left[ {{{\left( {\frac{3}{5}} \right)}^2}} \right]^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{3}{5}} \right)^{10}} = {\left( {\frac{3}{5}} \right)^5}\end{array}\)

d) \({\left( {1 – \frac{1}{3}} \right)^8}:{\left( {\frac{4}{9}} \right)^3} = {\left( {\frac{3}{3} – \frac{1}{3}} \right)^8}:{\left( ({\frac{2}{3}})^2 \right)^3}\\= {\left( {\frac{2}{3}} \right)^8}:{\left( {\frac{2}{3}} \right)^6} = {\left( {\frac{2}{3}} \right)^{8-6}}\\= {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)