Bài 30 trang 19 SGK Toán 7 tập 1

Đề bài

Tìm \(x\), biết

\(\begin{gathered}
a)\,\,x:{\left( { - \frac{1}{2}} \right)^3} = - \frac{1}{2} \hfill \\
b)\,\,\,\,{\left( {\,\frac{3}{4}} \right)^5}.x = {\left( {\frac{3}{4}} \right)^7} \hfill \\
\end{gathered} \)

Lời giải chi tiết

\(\eqalign{
& a)\,\,x:{\left( { - {1 \over 2}} \right)^3} = - {1 \over 2} \cr
& \,\,\,\,\,\,\,\,x = \left( { - {1 \over 2}} \right).{\left( { - {1 \over 2}} \right)^3} \cr & \,\,\,\,\,\,\,\,\,x = {\left( { - {1 \over 2}} \right)^{1+3}} \cr
& \,\,\,\,\,\,\,\,\,x = {\left( { - {1 \over 2}} \right)^4} \cr
& \,\,\,\,\,\,\,\,\,x = {1 \over {16}} \cr} \)

Vậy \(x = \frac{1}{16}\)

b) 

\(\begin{array}{l}{(\frac{3}{4})^5}.x = {(\frac{3}{4})^7}\\x = {(\frac{3}{4})^7}:{(\frac{3}{4})^5}\\x = {(\frac{3}{4})^{7 - 5}}\\x = {(\frac{3}{4})^2}\\x = \frac{9}{{16}}\end{array}\)

Vậy x = \(\frac{9}{16}\)

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