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9 tháng 9 2018

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)

\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)

\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)

\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)

\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)

\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)

\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)

\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)

\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)

7 tháng 9 2016

\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)

7 tháng 9 2016

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(S=\frac{4}{9}-\frac{1}{5}\)

\(S=\frac{11}{45}\)

14 tháng 9 2017

a) \(\frac{1}{3}-\left(\frac{1}{2}+\frac{1}{8}\right)\)

=   \(\frac{1}{3}-\left(\frac{4}{8}+\frac{1}{8}\right)\)

=     \(\frac{1}{3}-\frac{5}{8}\)

\(\frac{8}{24}-\frac{15}{24}\)

\(\frac{-7}{24}\)

b) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{13}+\frac{1}{8}\)

\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)\)\(\frac{1}{13}\)

\(\left(\frac{4}{8}-\frac{2}{8}+\frac{1}{8}\right)+\frac{1}{13}\)

=                 \(\frac{1}{8}+\frac{1}{13}\)

=                 \(\frac{13}{104}+\frac{8}{104}\)

=                        \(\frac{23}{104}\)

c) \(13\frac{2}{7}:\left(\frac{-8}{9}\right)+2\frac{5}{7}:\left(\frac{-8}{9}\right)\)

\(\left(13\frac{2}{7}+2\frac{5}{7}\right):\left(\frac{-8}{9}\right)\)

=         \(16:\left(\frac{-8}{9}\right)\)

=         -18