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)

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

Vậy \(x = \frac{1}{{16}}\).

 b)

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

Vậy \(x = \frac{9}{{25}}\).

c)

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

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).