Giải Bài 4 trang 15 sách bài tập toán 7 tập 1 - Chân trời sáng tạo

Đề bài

Tìm x, biết:

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

b) \(x.{\left( {\dfrac{{ - 3}}{7}} \right)^5} = {\left( {\dfrac{{ - 3}}{7}} \right)^7}\)

c) \({\left( {\dfrac{{ - 2}}{3}} \right)^{12}}:x = {\left( {\dfrac{{ - 2}}{3}} \right)^9}\)

d) \({\left( {x + \dfrac{1}{3}} \right)^2} = \dfrac{1}{{25}}\)

Lời giải chi tiết

\(\begin{array}{l}a)\,x:{\left( {\dfrac{{ - 1}}{3}} \right)^3} = \dfrac{{ - 1}}{3}\\ \Leftrightarrow x = \dfrac{{ - 1}}{3}.{\left( {\dfrac{{ - 1}}{3}} \right)^3}\\ \Leftrightarrow x = {\left( {\dfrac{{ - 1}}{3}} \right)^{ 4}}\\ \Leftrightarrow x = \dfrac{{( - 1).( - 1).( - 1).( - 1)}}{{3.3.3.3}} = \dfrac{1}{{81}}\end{array}\)

Vậy \(x=\dfrac{1}{81}\)

\(\begin{array}{l}b)\,x.{\left( {\dfrac{{ - 3}}{7}} \right)^5} = {\left( {\dfrac{{ - 3}}{7}} \right)^7}\\ \Leftrightarrow x = {\left( {\dfrac{{ - 3}}{7}} \right)^7}:{\left( {\dfrac{{ - 3}}{7}} \right)^5}\\ \Leftrightarrow x = {\left( {\dfrac{{ - 3}}{7}} \right)^{7 - 5}}\\ \Leftrightarrow x = {\left( {\dfrac{{ - 3}}{7}} \right)^2} = \dfrac{{( - 3).( - 3)}}{{7.7}}\\ \Leftrightarrow x = \dfrac{9}{{49}}\end{array}\)

Vậy \(x=\dfrac{9}{49}\)

\(\begin{array}{l}c)\,{\left( {\dfrac{{ - 2}}{3}} \right)^{12}}:x = {\left( {\dfrac{{ - 2}}{3}} \right)^9}\\ \Leftrightarrow {\left( {\dfrac{{ - 2}}{3}} \right)^{12}}:{\left( {\dfrac{{ - 2}}{3}} \right)^9} = x\\ \Leftrightarrow x = {\left( {\dfrac{{ - 2}}{3}} \right)^{12 - 9}}\\ \Leftrightarrow x = {\left( {\dfrac{{ - 2}}{3}} \right)^3}\\ \Leftrightarrow x = \dfrac{{( - 2).( - 2).( - 2)}}{{3.3.3}} = \dfrac{{ - 8}}{{27}}\end{array}\)

Vậy \(x=\dfrac{-8}{27}\)

\(\begin{array}{l}d){\left( {x + \dfrac{1}{3}} \right)^2} = \dfrac{1}{{25}}\\ \Leftrightarrow {\left( {x + \dfrac{1}{3}} \right)^2} = {\left( {\dfrac{1}{5}} \right)^2}\\TH1:x + \dfrac{1}{3} = \dfrac{1}{5}\\ \Leftrightarrow x + \dfrac{1}{3} = \dfrac{1}{5}\\ \Leftrightarrow x = \dfrac{1}{5} - \dfrac{1}{3} = \dfrac{{ - 2}}{{15}}\\TH2:x + \dfrac{1}{3} =  - \dfrac{1}{5}\\ \Leftrightarrow x + \dfrac{1}{3} =  - \dfrac{1}{5}\\ \Leftrightarrow x =  - \dfrac{1}{5} - \dfrac{1}{3} \\\Leftrightarrow x = \dfrac{-3}{15} - \dfrac{5}{15}\\\Leftrightarrow x = \dfrac{{ - 8}}{{15}}\end{array}\)

Vậy \(x\in\){\(\dfrac{-2}{15};\dfrac{-8}{15}\)}