Bài 6 (Trang 21, SGK Toán 7, Tập 1, Bộ Chân Trời Sáng Tạo)

Tính: 

$$\begin{gathered} a) [ {\left(\frac{3}{7}\right)}^4. {\left(\frac{3}{7}\right)}^5 ] : {\left(\frac{3}{7}\right)}^7; \\ b) [ {\left(\frac{7}{8}\right)}^5 : {\left(\frac{7}{8}\right)}^4 ].\frac{7}{8}; \\ c) [ {(0,6)}^3.{(0,6)}^8] : [{(0,6)}^7 .{(0,6)}^2] \end{gathered}$$ 

Hướng dẫn giải:

$$\begin{gathered} a) [ {\left(\frac{3}{7}\right)}^4 .{\left(\frac{3}{7}\right)}^5 ] :{\left(\frac{3}{7}\right)}^7 =[ {\left(\frac{3}{7}\right)}^{4+5}] :{\left(\frac{3}{7}\right)}^7 \\ ={ \left(\frac{3}{7}\right)}^9:{\left(\frac{3}{7}\right)}^7 \\ ={ \left(\frac{3}{7}\right)}^{9-7} \\ = {\left(\frac{3}{7}\right)}^2 \\ b) [ {\left(\frac{7}{8}\right)}^5: {\left(\frac{7}{8}\right)}^4 ].\frac{7}{8} \\ = \left[{\left(\frac{7}{8}\right)}^{5-4}\right] . \frac{7}{8} \\ ={ \left(\frac{7}{8}\right)}^1 .{\left(\frac{7}{8}\right)}^1 \\ ={ \left(\frac{7}{8}\right)}^{1+1} \\ ={ \left(\frac{7}{8}\right)}^2 \\ c) [{(0,6)}^3. {(0,6)}^8] : [{(0,6)}^7.{(0,6)}^2] \\ = [{(0,6)}^{3+8}] :[{(0,6)}^{7+2}] \\ = {(0,6)}^{11}: {(0,6)}^9 ={ (0,6)}^{11-9} \\ ={ (0,6)}^2 \end{gathered}$$