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\(\dfrac{81}{125}-\dfrac{8}{27}=\dfrac{3^4}{5^3}-\dfrac{2^3}{3^3}\)
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a) \(\left(\frac{27}{64}\right)^8:\left(\frac{3}{4}\right)^{22}=\left[\left(\frac{3}{4}\right)^3\right]^8:\left(\frac{3}{4}\right)^{22}=\left(\frac{3}{4}\right)^{24}:\left(\frac{3}{4}\right)^{22}=\left(\frac{3}{4}\right)^2\)
b) \(\left(\frac{2}{3}\right)^2.\left(-\frac{8}{27}\right).\left(-\frac{2}{3}\right)=\left(\frac{2}{3}\right)^2.\left(-\frac{2}{3}\right)^3.\left(-\frac{2}{3}\right)=\left(\frac{2}{3}\right)^2.\left(-\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^6\)
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Ta có:
\(\begin{array}{l}{\left( {\frac{1}{9}} \right)^5} = {[{\left( {\frac{1}{3}} \right)^2}]^5} = {(\frac{1}{3})^{2.5}} = {(\frac{1}{3})^{10}};\\{\left( {\frac{1}{{27}}} \right)^7} = {[{(\frac{1}{3})^3}]^7} = {(\frac{1}{3})^{3.7}} = {(\frac{1}{3})^{21}}\end{array}\)
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1. a. 2.8.16=16.16=\(16^2\)
b. 25.5.125=125.125=\(125^2\)
c. \(\frac{2}{3}.\frac{4}{9}.\frac{8}{27}=\frac{8}{27}.\frac{8}{27}=\left(\frac{8}{27}\right)^2\)
k cho em nha, em nhớ là em bày chị rồi mà
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`@` `\text {Ans}`
`\downarrow`
`-1/27 = ( (-1)^3/(-3)^3) =(-1/3)^3`
`8/729 = ( (2^3)/(9^3) ) = (2/9)^3`
`16/685 = ( (4^2)/( \sqrt {685}) ) = (4/(\sqrt {685}) )^2`
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\(2^6\)\(0,5^2\)\(\left(\frac{1}{2}\right)^4\)\(\left(\frac{1}{2}\right)^8\)\(\left(\frac{11}{12}\right)^2\)
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-8/27 = (-2/3)3 nha còn cái kia mình cụng ko bít đâu mình cũng đang hỏi
d)\(-\frac{8}{27}=\frac{\left(-2\right)^3}{3^3}=\left(-\frac{2}{3}\right)^3\)
h)\(-\frac{27}{64}=\frac{\left(-3\right)^3}{4^3}=\left(-\frac{3}{4}\right)^3\)
\(\frac{-8}{27}=\frac{\left(-2\right)^3}{3^3}\)
\(\frac{-27}{64}=\frac{-3^3}{4^3}\)