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\(3^2.\frac{1}{243.81^2}.\frac{1}{3^2}=\frac{1}{3^5.\left(3^4\right)^2}.\frac{3^2}{3^2}=\frac{1}{3^5.3^8}.1=\frac{1}{3^{13}}\)
Bài làm
\(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}\)
\(=\left(3^2.\frac{1}{3^2}\right).\left(\frac{1}{243}.81^2\right)\)
\(=1.\left(\frac{1}{243}.6561\right)\)
\(=27\)
# Học tốt #
3 mũ 2.1/243.81 mũ 2.1/3 mũ 2
= 3 mũ 2.1/3 mũ 5.(3 mũ 4)mũ 2.1/3 mũ 2
= 3 mũ 2.1/3 mũ 5.3 mũ 8.1/3 mũ 2
= 3 mũ 2.1/3 mũ 13.1/3 mũ 2
= 3 mũ 2/3 mũ 13.1/3 mũ 2
= 1/3 mũ 13=1/1594323
a) \(\left[\frac{1}{3}\right]^{50}.\left(-9\right)^{25}-\frac{2}{3}:4\)
\(\Rightarrow\frac{1}{3^{50}}.\left(-9\right)^{25}-\frac{2}{3}.\frac{1}{4}\)
\(\Rightarrow\frac{\left(-9\right)}{9^{25}}-\frac{1}{6}\)
\(\Rightarrow1-\frac{1}{6}\)
\(\Rightarrow\frac{6}{6}-\frac{1}{6}=\frac{5}{6}\)
Vậy = 5/6
a) \(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
\(=1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
=> 7S = \(7+\left(-1\right)+\left(-\frac{1}{7}\right)+...+\left(-\frac{1}{7}\right)^{2006}\)
Lấy 7S trừ S ta có :
7S - S = \(7+\left(-1\right)+\left(-\frac{1}{7}\right)+...+\left(-\frac{1}{7}\right)^{2006}-\left[1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\right]\)
6S = \(7-1-1+\left(\frac{1}{7}\right)^{2007}=5+\left(\frac{1}{7}\right)^{2007}\Rightarrow S=\frac{5+\left(\frac{1}{7}\right)^{2007}}{6}\)
\(\frac{2^3.5.7.5^2.7^3}{\left(2.5.7^2\right)^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=2.5=10\)
\(\frac{2^3\times5\times7\times5^2\times7^3}{\left(2\times5\times7^2\right)^2}=\frac{2^3\times5^3\times7^4}{2^2\times5^2\times7^4}=2\times5=10\)10
\(\left(\frac{1}{2}\right)^2\) . 2 = \(\frac{1}{4}\) .2 = \(\frac{2}{4}=\frac{1}{2}\)
\(\frac{1}{3}.\left(-2\right)^3-\frac{1}{3}=\frac{1}{3}.\left(-8\right)-\frac{1}{3}=\frac{-8}{3}-\frac{1}{3}=\frac{-9}{3}=-3\)
(1/4)×2=1/2
1/3×(-8)-1/3=1/3×(-9)=-3