Bài 8 (Trang 21, SGK Toán 7, Tập 1, Bộ Chân Trời Sáng Tạo)

Tính giá trị các biểu thức:

$$\begin{gathered} a) \frac{4^3.9^7}{{27}^5.8^2}; \\ b) \frac{{(-2)}^3.{(-2)}^7}{3.4^6}; \\ c) \frac{{(0,2)}^5.{(0,09)}^3}{{(0,2)}^5.{(0,3)}^4}; \\ d) \frac{2^3+2^4+2^5}{7^2}; \end{gathered}$$ 

Hướng dẫn giải:

$$\begin{gathered} a) \frac{4^3.9^7}{{27}^5.8^2}=\frac{{\left(2^2\right)}^3.{\left(3^2\right)}^7}{{\left(3^3\right)}^5.{\left(2^3\right)}^2} = \frac{2^6.3^{14}}{3^{15}.2^6}=\frac{1}{3} \\ b) \frac{{(-2)}^3.{(-2)}^7}{3.4^6}=\frac{{(-2)}^{3+7}}{3.{\left(2^2\right)}^6} \\ =\frac{{(-2)}^{10}}{3.2^{12}} = \frac{2^{10}}{3.2^{12}}=\frac{1}{3.2^2}=\frac{1}{12} \\ c) \frac{{(0,2)}^5.{(0,09)}^3}{{(0,2)}^7.{(0,3)}^4}=\frac{{(0,2)}^5.{(0,3^2)}^3}{{(0,2)}^7.{(0,3)}^4} \\ =\frac{{(0,2)}^5.{(0,3)}^6}{{(0,2)}^7.{(0,3)}^4}=\frac{0,3^2}{0,2^2}={\left(\frac{0,3}{0,2}\right)}^2 ={\left(\frac{3}{2}\right)}^2=\frac{9}{4} \\ d) \frac{2^3+ 2^4 + 2^5}{7^2}=\frac{8 + 16 + 32}{49}=\frac{56}{49}=\frac{8}{7} \end{gathered}$$