Giải bài 3 trang 20 SGK Toán 7 tập 1 - Chân trời sáng tạo

Đề bài

Tìm x, biết:

a) \(x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2};\)

b) \(x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9};\)

c) \({\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9};\)

d) \(x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\)

Lời giải chi tiết

a) \(x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\)

\(\begin{array}{l}x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b) \(x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\)

\(\begin{array}{l}x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c) \({\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\)

\(\begin{array}{l}x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d) \(x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\)

\(\begin{array}{l}x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).