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\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{3}{20\cdot23}\)
\(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{23}=\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
\(\frac{2}{5\cdot7}+\frac{5}{7\cdot12}+\frac{8}{12\cdot20}+\frac{3}{140}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{20}+\frac{3}{140}\)
\(=\frac{1}{5}-\frac{1}{20}+\frac{3}{140}\)
\(=\frac{3}{20}+\frac{3}{140}=\frac{6}{35}\)
\(\frac{2}{5.7}\)+ \(\frac{5}{7.12}\)+ \(\frac{8}{12.20}\)+ \(\frac{3}{140}\).
= \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{12}\)+ \(\frac{1}{12}\)- \(\frac{1}{20}\)+ \(\frac{3}{140}\).
= \(\frac{1}{5}\)- \(\frac{1}{20}\)+ \(\frac{3}{140}\).
= \(\frac{20}{100}\)- \(\frac{5}{100}\)+ \(\frac{3}{140}\).
= \(\frac{15}{100}\)+ \(\frac{3}{140}\).
= \(\frac{3}{20}\)+ \(\frac{3}{140}\).
= \(\frac{21}{140}\)+ \(\frac{3}{140}\).
= \(\frac{24}{140}\).
= \(\frac{6}{35}\).
a) \(C=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(C=1.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
\(C=1.\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(C=1.\frac{9}{38}\)
\(C=\frac{9}{38}\)
b) \(D=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{99.100}\)
\(D=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\right)\)
\(D=5.\left(\frac{1}{10}-\frac{1}{100}\right)\)
\(D=5.\frac{9}{100}\)
\(D=\frac{99}{20}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\frac{1}{10.11}+7.\frac{1}{11.12}+7.\frac{1}{12.13}+...+7.\frac{1}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=\frac{6}{70}\)
\(=\frac{3}{35}\)
\(\frac{4}{3\cdot7}+\frac{5}{7.12}+..+\frac{8}{20\cdot28}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{28}\)
\(=\frac{1}{3}-\frac{1}{28}+0+...+0\)
\(=\frac{25}{84}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{28}\)
\(=\frac{1}{3}-\frac{1}{28}\)
\(=\frac{25}{84}\)