Bài 5 trang 15 SBT Toán 7 tập 1 Chân trời sáng tạo: Tính...

Tính

a) $\left[ {{{\left( {\dfrac{2}{5}} \right)}^6}.{{\left( {\dfrac{2}{5}} \right)}^5}} \right]:{\left( {\dfrac{2}{5}} \right)^9}$

b) $\left[ {{{\left( {\dfrac{3}{7}} \right)}^8}:{{\left( {\dfrac{3}{7}} \right)}^7}} \right].\left( {\dfrac{3}{7}} \right)$

c) $\left[ {{{\left( {\dfrac{2}{5}} \right)}^9}.{{\left( {\dfrac{2}{5}} \right)}^4}} \right]:\left[ {{{\left( {\dfrac{2}{5}} \right)}^7}.{{\left( {\dfrac{2}{5}} \right)}^3}} \right]$ 

Nhân, chia 2 lũy thừa có cùng cơ số

$\begin{array}{l}a)\left[ {{{\left( {\dfrac{2}{5}} \right)}^6}.{{\left( {\dfrac{2}{5}} \right)}^5}} \right]:{\left( {\dfrac{2}{5}} \right)^9} = {\left( {\dfrac{2}{5}} \right)^{5 + 6}}:{\left( {\dfrac{2}{5}} \right)^9} = {\left( {\dfrac{2}{5}} \right)^{11 - 9}} = \dfrac{{2.2}}{{5.5}} = \dfrac{4}{{25}}\\b)\left[ {{{\left( {\dfrac{3}{7}} \right)}^8}:{{\left( {\dfrac{3}{7}} \right)}^7}} \right].\left( {\dfrac{3}{7}} \right) = \left( {\dfrac{3}{7}} \right).\left( {\dfrac{3}{7}} \right) = \dfrac{9}{{49}}\\c)\left[ {{{\left( {\dfrac{2}{5}} \right)}^9}.{{\left( {\dfrac{2}{5}} \right)}^4}} \right]:\left[ {{{\left( {\dfrac{2}{5}} \right)}^7}.{{\left( {\dfrac{2}{5}} \right)}^3}} \right] = {\left( {\dfrac{2}{5}} \right)^{13}}:{\left( {\dfrac{2}{5}} \right)^{10}} = {\left( {\dfrac{2}{5}} \right)^3} = \dfrac{{2.2.2}}{{5.5.5}} = \dfrac{8}{{125}}\end{array}$