Giải bài 9 trang 16 vở thực hành Toán 7

Đề bài

Tính \(A = \left( {{{\left( {\frac{1}{{125}}} \right)}^3} + {{\left( {\frac{1}{5}} \right)}^5}} \right):\left( {{{\left( {\frac{1}{5}} \right)}^5} + \frac{1}{5}} \right).\)

Lời giải chi tiết

Ta có:

\(\begin{array}{l}{\left( {\frac{1}{{125}}} \right)^3} + {\left( {\frac{1}{5}} \right)^5} = {\left[ {{{\left( {\frac{1}{5}} \right)}^3}} \right]^3} + {\left( {\frac{1}{5}} \right)^5}\\ = {\left( {\frac{1}{5}} \right)^9} + {\left( {\frac{1}{5}} \right)^5} = {\left( {\frac{1}{5}} \right)^5}\left[ {{{\left( {\frac{1}{5}} \right)}^4} + 1} \right]\end{array}\)

\({\left( {\frac{1}{5}} \right)^5} + \frac{1}{5} = \frac{1}{5}.\left[ {{{\left( {\frac{1}{5}} \right)}^4} + 1} \right]\)

Vậy

\(\begin{array}{l}A = {\left( {\frac{1}{5}} \right)^5}\left[ {{{\left( {\frac{1}{5}} \right)}^4} + 1} \right]:\left\{ {\left( {\frac{1}{5}} \right).\left[ {{{\left( {\frac{1}{5}} \right)}^4} + 1} \right]} \right\}\\ = {\left( {\frac{1}{5}} \right)^5}:\left( {\frac{1}{5}} \right) = {\left( {\frac{1}{5}} \right)^4}.\end{array}\)