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6 tháng 4 2017

\(A=1.\left(\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=\frac{1}{4}-\frac{1}{50}=\frac{23}{100}\)

mk nhanh nhất

6 tháng 4 2017

= 23 /100 chac 100 % lun

4 tháng 8 2016

a)A=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

      =\(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

      =\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

      =\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

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

      =\(\frac{-98}{100}=\frac{-49}{50}\)

4 tháng 8 2016

a) A\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\) 

A=\(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)( vận dụng quy tắc dấu ngoặc)\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

 \(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\) = \(\frac{1}{100}-\frac{99}{100}\) = \(\frac{-98}{100}\) = \(\frac{-49}{50}\)

13 tháng 5 2019

Ta co:

\(\frac{1}{2}A=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}+\frac{1}{90}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}+\frac{1}{9.10}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{10}=\frac{13}{20}\Rightarrow A=\frac{13}{10}.\)

13 tháng 5 2019

\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{36}+\frac{1}{45}\)

\(A=\frac{2}{4}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{72}+\frac{2}{90}\)

\(A=\frac{2}{2.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{8.9}+\frac{2}{9.10}\)

\(A=2\left(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

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

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

\(A=2.\frac{2}{5}\)

\(A=\frac{4}{5}\)

~ Học tốt ~ K cho mk nhé! Thank you.

7 tháng 8 2016

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)

\(A=\frac{1}{6}-\frac{1}{15}\)

\(A=\frac{1}{10}\)

7 tháng 8 2016

A=\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

=\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

=\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{14}-\frac{1}{15}\)

=\(\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

20 tháng 7 2016

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+...+\frac{1}{210}=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{14.15}\)

\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{14}-\frac{1}{15}\)

\(=\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

20 tháng 7 2016

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)

\(A=\frac{1}{6}-\frac{1}{15}\)

\(A=\frac{1}{10}\)

19 tháng 4 2016

\(A=-\frac{1}{20}+-\frac{1}{30}+...+-\frac{1}{90}\)

   \(=-\frac{1}{4.5}+-\frac{1}{5.6}+...+-\frac{1}{9.10}\)

   \(=\left(-\frac{1}{4}\right)-\left(-\frac{1}{5}\right)+\left(-\frac{1}{5}\right)-\left(-\frac{1}{6}\right)+...+\left(-\frac{1}{9}\right)-\left(-\frac{1}{10}\right)\)

   \(=\left(-\frac{1}{4}\right)-\left(-\frac{1}{10}\right)=-\frac{3}{20}\)

Vậy \(A=-\frac{3}{20}\)

 

2 tháng 2 2016

1/

A= 1/15+1/35+1/63+1/99+ ... + 1/9999

A=1/3.5+1/5.7+1/7.9+ ... +1/99.101

2A=2/3.5+2/5.7+2/7.9+ ... +2/99.101

2A=1/3-1/5+1/5-1/7+1/7-1/9+ ... + 1/99-1/101

2A=1/3-1/101

A=49/303

Sai thì thôi nhé

2 tháng 2 2016

A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

A=1-1/7

A=6/7

22 tháng 10 2018

mk làm được rồi G=4

18 tháng 5 2016

1/4,5  + 1/3 

= 2/9  +1/3

=2/3 +3/9

= 5/9

 

18 tháng 5 2016

1/4,5+1/3

=2/9+1/3

=2/9+3/9

=5/9