Giải bài 6 trang 21

Đề bài

Tính:

a)$\left[{\left(\frac{3}{7}\right)}^4.{\left(\frac{3}{7}\right)}^5\right]:{\left(\frac{3}{7}\right)}^7;$          

b)$\left[{\left(\frac{7}{8}\right)}^5:{\left(\frac{7}{8}\right)}^4\right].\left(\frac{7}{8}\right);$

c)$\left[{\left(0,6\right)}^3.{\left(0,6\right)}^8\right]:\left[{\left(0,6\right)}^7.{\left(0,6\right)}^2\right]$.

Lời giải chi tiết

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