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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) \(\left(\frac{2}{3}\right)^2.\left(-\frac{8}{27}\right).\left(-\frac{2}{3}\right)=\left(\frac{2}{3}\right)^2.\left(-\frac{2}{3}\right)^3.\left(-\frac{2}{3}\right)=\left(\frac{2}{3}\right)^2.\left(-\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^6\)
Ta có:
\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)
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
a)\({\left[ {{{\left( { - \frac{1}{6}} \right)}^3}} \right]^4}\) (với \(a = - \frac{1}{6}\))
\(=(- \frac{1}{6})^{3. 4}=(- \frac{1}{6})^{12}\)
b)\({\left[ {{{\left( { - 0,2} \right)}^4}} \right]^5}\) (với \(a = - 0,2\))
\(=(-0,2)^{4.5}=(-0,2)^{20}\)
\(\frac{8^{11}.3^{17}}{27^{10}.9^{15}}=\frac{8^{11}.3^{17}}{3^{30}.3^{30}}=\frac{8^{11}}{3^{13}.3^{30}}=\frac{8^{11}}{3^{43}}\)
\(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{[\left(5-1\right).5^3]^3}{5^{12}}=\frac{\left(4.5^3\right)^3}{5^{12}}=\frac{64.5^9}{5^{12}}=\frac{64}{5^3}=\left(\frac{4}{5}\right)^3\)
\(\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+6^{20}}{6^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{30}\right)}{3^{20}.\left(2^{20}-2+3^{20}\right)}=\frac{2^{20}}{3^{20}}=\left(\frac{2}{3}\right)^{20}\)
\(\left(\frac{1}{8}\right)^4:\left(-\frac{1}{4}\right)^5\)
\(=\frac{1}{8^4}:\frac{-1}{4^5}\)
\(=\frac{1}{2^{12}}:\frac{-1}{2^{10}}\)
\(=\frac{1}{2^{12}}.\frac{2^{10}}{-1}=\frac{1}{-4}\)