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27 tháng 3 2017

\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90} \)\(\frac{-1}{90}\)

\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)

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

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

27 tháng 3 2017

\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

= \(\frac{-1}{4}-\frac{-1}{10}\)

=\(\frac{-3}{20}\)

23 tháng 1 2020

\(a,A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)

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

\(=-\frac{1}{4}+\frac{1}{10}\)

\(=-\frac{3}{20}\)

\(b,B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{11.7}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-....-\frac{1}{28}\)

\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

23 tháng 1 2020

a) \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

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

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

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{9}-\frac{1}{10}=\frac{1}{4}-\frac{1}{10}=\frac{3}{20}\)

\(\Rightarrow A=\frac{3}{20}:\left(-1\right)=\frac{-3}{20}\)

b) \(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(\Rightarrow\frac{1}{7}B=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

\(\Rightarrow B=\frac{13}{28}:\frac{1}{7}=\frac{13}{28}.7=\frac{13}{4}\)

19 tháng 7 2018

\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+...+-\frac{1}{90}\)

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

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

\(A=-\frac{3}{20}\)

3 tháng 2 2016

\(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(A=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)

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

\(A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)=-\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{-3}{20}\)

5 tháng 3 2016

a) A = \(\frac{1}{5}\) - \(\frac{1}{4}\)\(\frac{1}{6}\)\(\frac{1}{5}\)\(\frac{1}{7}\)-\(\frac{1}{6}\)+\(\frac{1}{8}\)-\(\frac{1}{7}\)+\(\frac{1}{9}\)\(\frac{1}{8}\)\(\frac{1}{10}\)\(\frac{1}{9}\)

    = \(\frac{-1}{4}\)+\(\frac{1}{10}\)\(\frac{-6}{40}\)\(\frac{-3}{20}\)

b) B = \(\frac{5}{2.1}\)+ \(\frac{1}{11}\)(4 + \(\frac{3}{2}\)) + \(\frac{1}{2.15}\)(1 + \(\frac{13}{2}\))

        = \(\frac{5}{2.1}\)\(\frac{1}{11}\).\(\frac{11}{2}\)\(\frac{1}{2.15}\).\(\frac{15}{2}\)

        = \(\frac{5}{2}\)\(\frac{1}{2}\)\(\frac{1}{4}\)= 3 + \(\frac{1}{4}\)\(\frac{13}{4}\)

19 tháng 11 2021

\(a,\Rightarrow A=-1\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{9.10}\right)\)

\(\Rightarrow A=-1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(\Rightarrow A=-1\left(\dfrac{1}{4}-\dfrac{1}{10}\right)\)

\(\Rightarrow A=\dfrac{-3}{20}\)

 

19 tháng 11 2021

Bài 2:

\(a,\dfrac{1717}{8585}=\dfrac{1717:1717}{8585:1717}=\dfrac{1}{5};\dfrac{1313}{5151}=\dfrac{1313:101}{5151:101}=\dfrac{13}{51}\\ \dfrac{1}{5}=\dfrac{51}{255}< \dfrac{65}{255}=\dfrac{13}{51}\\ \Rightarrow\dfrac{1717}{8585}< \dfrac{1313}{5151}\)

\(b,\dfrac{201201}{202202}=\dfrac{201201:1001}{202202:1001}=\dfrac{201}{202}=\dfrac{201\cdot1001001}{202\cdot1001001}=\dfrac{201201201}{202202202}\)