Bài 7 (Trang 21, SGK Toán 7, Tập 1, Bộ Chân Trời Sáng Tạo)

Tính:

$$\begin{gathered} a) {(\frac{2}{5}+\frac{1}{2})}^2; \\ b) {(0,75-1\frac{1}{2})}^3; \\ c) {(1-\frac{1}{3})}^8 :{\left(\frac{4}{9}\right)}^3; \end{gathered}$$

Hướng dẫn giải:

$$\begin{gathered} a) {(\frac{2}{5}+\frac{1}{2})}^2=\hspace{0.17em}{(\frac{4}{10}+\frac{5}{10})}^2 = {\left(\frac{9}{10}\right)}^2 = \frac{9^2}{{10}^2} =\frac{81}{100} \\ b) {(0,75-1\frac{1}{2})}^3 ={(\frac{3}{4}-\frac{3}{2})}^3 ={(\frac{3}{4}-\frac{6}{4})}^3 ={\left(\frac{-3}{4}\right)}^3 =\frac{{(-3)}^3}{4^3}=\frac{-27}{64} \\ c) {\left(\frac{3}{5}\right)}^{15} :{(0,36)}^5 =\frac{3^{15}}{5^{15}}:{\left(\frac{9}{25}\right)}^5 =\frac{3^{15}}{5^{15}}:{\left(\frac{3^2}{5^2}\right)}^5 =\frac{3^{15}}{5^{15}}:\frac{3^{10}}{5^{10}} \\ =\frac{3^{15}}{5^{15}}.\frac{5^{10}}{3^{10}}=\frac{3^{15}.5^{10}}{5^{15}.3^{10}} \\ =\frac{3^5}{5^5}={ \left(\frac{3}{5}\right)}^5 \\ d) {(1-\frac{1}{3})}^8:{\left(\frac{4}{9}\right)}^3 ={ \left(\frac{2}{3}\right)}^8 :[\hspace{0.17em}{\left(\frac{2}{3}\right)}^2{]}^3 \\ ={\left(\frac{2}{3}\right)}^8 : {\left(\frac{2}{3}\right)}^{2.3} ={ \left(\frac{2}{3}\right)}^8 :{\left(\frac{2}{3}\right)}^6 ={ \left(\frac{2}{3}\right)}^{8-6} ={ \left(\frac{2}{3}\right)}^2 \end{gathered}$$