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10 tháng 10 2016

\(G=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)

\(G=2.\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{504.509}\right)\)

\(G=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)

\(G=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)

\(G=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)

\(G=\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)

10 tháng 10 2016

Khó quá !!!

27 tháng 6 2017

Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)

=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)

=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)

=\(\frac{1}{2}-\frac{1}{509}\)

=\(\frac{507}{1018}\)

MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))

27 tháng 6 2017

làm 2 câu còn lại đi câu đó làm rồi

11 tháng 8 2020

a) \(\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{202.509}=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)

\(=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)

\(=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)

\(=\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)

b) \(\frac{1}{10.9}+\frac{1}{18.13}+...+\frac{1}{802.405}=\frac{2}{10.18}+\frac{2}{18.26}+...+\frac{2}{802.810}\)

\(=\frac{2}{8}\left(\frac{8}{10.18}+\frac{8}{18.26}+...+\frac{8}{802.810}\right)=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+...+\frac{1}{802}-\frac{1}{810}\right)\)

\(=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{810}\right)=\frac{1}{4}.\frac{40}{405}=\frac{10}{405}\)

11 tháng 8 2020

Bạn vào câu hỏi tương tự tham khảo !

22 tháng 5 2015

\(\frac{654}{12254}=\frac{12254-11600}{12254}=1+\frac{-11600}{12254}=1+\frac{1}{\frac{12254}{-11600}}=1+\frac{1}{1+\frac{23854}{-11600}}=1+\frac{1}{1+\frac{1}{-\frac{11600}{23854}}}=\)sức gõ công thức có hạn, cứ theo đó mà làm tiếp, đảm bảo sẽ ra ngay kết quả

đúng nha bạn

7 tháng 2 2020

\(G=\left(\frac{1}{3}-1\right)\left(\frac{1}{6}-1\right)\left(\frac{1}{15}-1\right)\left(\frac{1}{21}-1\right)\left(\frac{1}{36}-1\right)\)

\(\Leftrightarrow G=-\frac{2}{3}.\frac{-5}{6}.\frac{-14}{15}.\frac{-20}{21}.\frac{-35}{36}\)

\(\Leftrightarrow G=\frac{-2.}{3}.\frac{5}{6}.\frac{14}{15}.\frac{20}{21}.\frac{35}{36}\)

\(\Leftrightarrow G=\frac{-2.5.2.7.2.2.5.5.7}{3.2.3.3.5.3.7.2.3.2.3}\)

\(\Leftrightarrow G=\frac{-2^4.5^3.7^2}{2^3.3^6.5.7}\)

\(\Leftrightarrow G=\frac{-2.5^2.7}{3^6}\)

\(\Leftrightarrow G=\frac{-350}{729}\)

P/s : Xin lỗi vì cách giải cùi bắp của mình :((

8 tháng 8 2016

\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)

  \(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)

     \(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)

\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)     \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)

  \(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)

\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)

   \(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'

    \(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)

\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)

     \(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)

     \(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)

        \(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)

\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)

    \(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)

      \(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)

        \(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)

         \(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)