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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
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:
\(\begin{array}{l}{\left( {\frac{1}{9}} \right)^5} = {[{\left( {\frac{1}{3}} \right)^2}]^5} = {(\frac{1}{3})^{2.5}} = {(\frac{1}{3})^{10}};\\{\left( {\frac{1}{{27}}} \right)^7} = {[{(\frac{1}{3})^3}]^7} = {(\frac{1}{3})^{3.7}} = {(\frac{1}{3})^{21}}\end{array}\)
d)\(-\frac{8}{27}=\frac{\left(-2\right)^3}{3^3}=\left(-\frac{2}{3}\right)^3\)
h)\(-\frac{27}{64}=\frac{\left(-3\right)^3}{4^3}=\left(-\frac{3}{4}\right)^3\)
\(\frac{-8}{27}=\frac{\left(-2\right)^3}{3^3}\)
\(\frac{-27}{64}=\frac{-3^3}{4^3}\)
\(\frac{8^{11}.3^{17}}{27^{10}.9^{15}}=\frac{8^{11}.3^{17}}{3^{30}.3^{30}}=\frac{8^{11}}{3^{13}.3^{30}}=\frac{8^{11}}{3^{43}}\)
\(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{[\left(5-1\right).5^3]^3}{5^{12}}=\frac{\left(4.5^3\right)^3}{5^{12}}=\frac{64.5^9}{5^{12}}=\frac{64}{5^3}=\left(\frac{4}{5}\right)^3\)
\(\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+6^{20}}{6^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{30}\right)}{3^{20}.\left(2^{20}-2+3^{20}\right)}=\frac{2^{20}}{3^{20}}=\left(\frac{2}{3}\right)^{20}\)
\(a,=\frac{2}{3}.\frac{2^2}{3^2}.\frac{2^3}{3^3}=\frac{2.2^2.2^3}{3.3^2.3^3}=\frac{2^6}{3^6}\)
\(b,=\frac{3}{4}.\frac{3^2}{4^2}.\frac{3^3}{4^3}=\frac{3.3^2.3^3}{4.4^2.4^3}=\frac{3^6}{4^4}\)
Mk nhầm chút nhé câu b\(=\frac{3^6}{4^6}\)