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\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\)
\(4B=\frac{44}{45}\)
\(B=\frac{11}{45}\)
\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{41.45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.2^{10}}=\frac{2^{2.5}.3^{2.4}-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^0}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^8.3^8}=\frac{2^{10}.3^8\left(1-3\right)}{2^8.3^8\left(2^2+1\right)}=\frac{4.-2}{5}=-\frac{8}{5}\)
Trả lời :
\(E=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{n\left(n+4\right)}\right)\)
\(\Rightarrow E=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n}-\frac{1}{n+4}\right)\)
\(\Rightarrow E=-\left(1-\frac{1}{n+4}\right)\)
\(\Rightarrow E=1+\frac{1}{n+4}\)
P/s : Sai thì thông cảm nha chị. Dạng này lâu chưa làm nên không nhớ rõ.
\(E=-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.11}-...-\frac{4}{\left(n-4\right)n}\)
\(\Rightarrow E=-\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.11}+...+\frac{4}{\left(n-4\right)n}\right)\)
\(\Rightarrow E=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(\Rightarrow E=-\left(1-\frac{1}{n}\right)\)
\(\Rightarrow E=-1+\frac{1}{n}\)
a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)
b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)
c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)
\(=\left(-4,1-5,9\right)+\left(-13,7-6,3\right)+59\\ =-10-20+59=29\)