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11 tháng 5

    \(\dfrac{1}{1.6}\) + \(\dfrac{1}{3.10}\) + \(\dfrac{1}{5.14}\) + ... + \(\dfrac{1}{31.66}\)

=   \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{3.2.5}\) + \(\dfrac{1}{5.2.7}\) + ... + \(\dfrac{1}{31.2.32}\)

\(\dfrac{1}{2}\).(\(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + ... + \(\dfrac{1}{31.32}\))

\(\dfrac{1}{2}\)\(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{31.32}\)

\(\dfrac{1}{2}\).\(\dfrac{1}{2}\)(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{31}\) - \(\dfrac{1}{32}\))

\(\dfrac{1}{2}\).\(\dfrac{1}{2}\)( 1 - \(\dfrac{1}{32}\))

\(\dfrac{1}{2}.\dfrac{1}{2}\).\(\dfrac{31}{32}\)

\(\dfrac{31}{128}\)

26 tháng 4 2018

\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)

=\(\dfrac{3+6+12+21\cdot35}{14+28+7\cdot45}\)

=\(\dfrac{450}{119}\)

\(\dfrac{450}{119}>1\)\(1>\dfrac{303}{708}\)

\(\Rightarrow\)\(\dfrac{450}{119}>\dfrac{303}{708}\)

2 tháng 7 2017

\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)

\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)

\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)

\(=\dfrac{1.3.5}{1.5.7}\)

2 tháng 7 2017

giúp mk vs các bn

11 tháng 4 2018

Giải:

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

Vậy ...

11 tháng 4 2018

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

=\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

=\(\dfrac{1}{3}-\dfrac{1}{10}\)

=\(\dfrac{7}{30}\)

23 tháng 7 2017

a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=5\left(1-\dfrac{1}{100}\right)\)

\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)

b, \(C=1.2.3+2.3.4+...+8.9.10\)

\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)

\(C=\dfrac{8.9.10.11}{4}=1980.\)

c, https://hoc24.vn/hoi-dap/question/384591.html

Câu này bạn vào đây mình đã giải câu tương tự nhé.

23 tháng 7 2017

\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{99}{20}\)

14 tháng 10

 

????

 

17 tháng 4 2022

\(\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

\(\dfrac{22}{77}-\dfrac{14}{77}=\dfrac{8}{77}\)

\(\dfrac{11}{13}\times\dfrac{26}{31}=11\times\dfrac{2}{31}=\dfrac{22}{31}\)

\(\dfrac{1}{2}\times3\times\dfrac{2}{5}=\dfrac{3}{5}\)

17 tháng 4 2022

mình cảm ơn

16 tháng 2 2022

1100444-88888=

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

\(\frac{10}{22}\)