Nguyễn Tuấn Kiệt

a, \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+....+\frac{1}{24\cdot25}\)

b, \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{99\cdot101}\)

c, \(5\frac{2}{7}\cdot\frac{8}{11}+5\frac{2}{7}\cdot\frac{5}{11}-5\frac{2}{7}\cdot\frac{2}{11}\)

giup minh voi mai nop roi!!!!

T.Ps
5 tháng 7 2019 lúc 20:03

#)Giải :

a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

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Xyz OLM
5 tháng 7 2019 lúc 20:08

a) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)

\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)

\(\frac{1}{5}-\frac{1}{25}\)

\(\frac{4}{25}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(1-\frac{1}{101}\)

\(\frac{100}{101}\)

c) \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)

\(5\frac{2}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)

\(5\frac{2}{7}\)

\(\frac{37}{7}\)

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a)\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

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a,\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b,\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}\)

\(=\frac{100}{101}\)

c,\(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)

\(=5\frac{2}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)

\(=\frac{37}{7}.1\)

\(=\frac{37}{7}\)

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