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\(1:243=\frac{1}{243}=\left(\frac{1}{3}\right)^5\)
\(1:3=\left(\frac{1}{3}\right)^1\)
\(\frac{1}{9}=\left(\frac{1}{3}\right)^3\)
1/81 = (1/9)^2
243 = 3^5
8. 3^3 = 2^3 . 3^3 = 6^3
81.2^8 = 9^2 . (2^4)^2 = 9^2 . 16^2 = ( 9. 16)^2 = 144^2
\(a,3^2\cdot\frac{1}{243}=3^2\cdot\frac{1}{3^5}=\frac{1}{3^3}=\frac{1^3}{3^3}=\left(\frac{1}{3}\right)^3\)
\(b,81^2\cdot\frac{1}{3^3}=\left(3^4\right)^2\cdot\frac{1}{3^3}=3^8\cdot\frac{1}{3^3}=3^5\)
a) \(3^2.\frac{1}{243}=\frac{1.3^3}{243}=\frac{3^2}{243}=\frac{3^2}{3^5}=\frac{1}{3^3}=\frac{1}{27}\)
b) \(81^2.\frac{1}{3^3}=\frac{1.81^2}{3^3}=\frac{81^2}{3^3}=\frac{3^8}{3^3}=3^5=243\)
Ta có:
\(\begin{array}{l}{\left( {\frac{1}{9}} \right)^5} = {[{\left( {\frac{1}{3}} \right)^2}]^5} = {(\frac{1}{3})^{2.5}} = {(\frac{1}{3})^{10}};\\{\left( {\frac{1}{{27}}} \right)^7} = {[{(\frac{1}{3})^3}]^7} = {(\frac{1}{3})^{3.7}} = {(\frac{1}{3})^{21}}\end{array}\)
\(25\cdot5^3\cdot\dfrac{1}{625}\cdot5^3=5^8\cdot\dfrac{1}{5^4}=5^4\)
\(1=3^0\)
\(243=3^5\)
\(\dfrac{1}{3}=3^{-1}\)
\(\dfrac{1}{9}=3^{-2}\)