\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{8}{20\cdot28}\)
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HA
\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{3}{20\cdot23}\)
1
10 tháng 8 2017
\(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{23}=\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
1 tháng 8 2018
\(=\frac{219}{520}=\frac{155052}{368160}\)
\(=\frac{303}{708}=\frac{157560}{368160}\)
\(\frac{155052}{368160}< \frac{157560}{368160}\)
VẬY \(\frac{303}{708}\)LỚN HƠN
29 tháng 7 2018
\(\frac{2}{5\cdot7}+\frac{5}{7\cdot12}+\frac{8}{12\cdot20}+\frac{3}{140}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{20}+\frac{3}{140}\)
\(=\frac{1}{5}-\frac{1}{20}+\frac{3}{140}\)
\(=\frac{3}{20}+\frac{3}{140}=\frac{6}{35}\)
29 tháng 7 2018
\(\frac{2}{5.7}\)+ \(\frac{5}{7.12}\)+ \(\frac{8}{12.20}\)+ \(\frac{3}{140}\).
= \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{12}\)+ \(\frac{1}{12}\)- \(\frac{1}{20}\)+ \(\frac{3}{140}\).
= \(\frac{1}{5}\)- \(\frac{1}{20}\)+ \(\frac{3}{140}\).
= \(\frac{20}{100}\)- \(\frac{5}{100}\)+ \(\frac{3}{140}\).
= \(\frac{15}{100}\)+ \(\frac{3}{140}\).
= \(\frac{3}{20}\)+ \(\frac{3}{140}\).
= \(\frac{21}{140}\)+ \(\frac{3}{140}\).
= \(\frac{24}{140}\).
= \(\frac{6}{35}\).
1 tháng 3 2019
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) \(D=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{99.100}\)
\(D=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\right)\)
\(D=5.\left(\frac{1}{10}-\frac{1}{100}\right)\)
\(D=5.\frac{9}{100}\)
\(D=\frac{99}{20}\)
\(\frac{4}{3\cdot7}+\frac{5}{7.12}+..+\frac{8}{20\cdot28}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{28}\)
\(=\frac{1}{3}-\frac{1}{28}+0+...+0\)
\(=\frac{25}{84}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{28}\)
\(=\frac{1}{3}-\frac{1}{28}\)
\(=\frac{25}{84}\)