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\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{\left(2n-1\right)\left(2n+1\right)}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=2.\left(1-\frac{1}{2n+1}\right)\)
\(=2.\left(\frac{2n}{2n+1}\right)\)
\(=\frac{4n}{2n+1}\)
Tham khảo nhé~
\(=\dfrac{-7^4\left(7\cdot5-1\right)}{7^6\cdot10}-\dfrac{27}{100}\)
\(=\dfrac{-34}{7^2\cdot10}-\dfrac{27}{100}\)
\(=\dfrac{-1663}{4900}\)
`#3107.101107`
\(\dfrac{27^2\cdot2^3\cdot5^4}{15^2\cdot6^9}\)
\(=\dfrac{\left(3^3\right)^2\cdot2^3\cdot5^4}{3^2\cdot5^2\cdot2^9\cdot3^9}\)
\(=\dfrac{3^6\cdot2^3\cdot5^4}{3^{11}\cdot5^2\cdot2^9}\)
\(=\dfrac{5^2}{3^5\cdot2^6}\)
\(=\dfrac{25}{15552}\)
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{97\cdot99}\)
\(=2\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(1-\frac{1}{99}\right)\)
\(=2\cdot\frac{98}{99}\)
\(=\frac{196}{99}\)