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3 tháng 6 2016

Ta có: \(M=\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-...-\frac{1}{1.2}\)

\(\Rightarrow-M=-\left(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-...-\frac{1}{1.2}\right)\)

\(\Rightarrow-M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}-\frac{1}{9.10}\)

\(\Rightarrow-M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}-\frac{1}{90}\)

\(\Rightarrow-M=1-\frac{1}{9}-\frac{1}{90}\)

\(\Rightarrow-M=\frac{8}{9}-\frac{1}{90}=\frac{80}{90}-\frac{1}{90}=\frac{79}{90}\)

\(\Rightarrow M=-\frac{79}{90}\)

15 tháng 7 2017

1/5.6  - 1/6/7 - 1/7.8 - ...- 1/2016.2017

= 1(1/5 - 1/6 +1/6-1/7+ 1/7 - 1/8 +...+1/2016 - 1/2017)

= 1(1/5 - 1/2017)

= 1.2012/10085

= 2012/10085

15 tháng 7 2017

1/5.6 - 1/6.7 - 1/7.8 - ... 1/2016.2017

= 1 ( 1/5 - 1/6 - 1/7 - 1/8 + ... + 1/2016 - 1/2017 )

= 1 ( 1/5 - 1/2017 )

= 1.2012/10085

= 2012/10085

3 tháng 7 2015

\(\frac{1}{5.6}\)\(\frac{1}{6.7}\)\(\frac{1}{7.8}\) - ... - \(\frac{1}{2004.2005}\) 

\(\frac{1}{5}\)\(\frac{1}{6}\)\(\frac{1}{6}\)\(\frac{1}{7}\)\(\frac{1}{7}\)\(\frac{1}{8}\)+ ... + \(\frac{1}{2004}\)\(\frac{1}{2005}\)

=\(\frac{1}{5}\)\(\frac{1}{2005}\) 

=  \(\frac{80}{401}\)

14 tháng 8 2016

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

\(B=\frac{1}{3}-\frac{1}{10}\)

\(B=\frac{7}{30}\)

14 tháng 8 2016

sai đề

15 tháng 7 2018

\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)+\frac{1}{9.10}\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)

\(=-\left(1-\frac{1}{10}\right)+\frac{1}{90}\)

\(=-\frac{9}{10}+\frac{1}{90}\)

= ...

bn tự tính nha!
 

22 tháng 10 2017

\(B=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)

\(\Rightarrow B=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(\Rightarrow B=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

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

\(\Rightarrow B=\frac{1}{12}-\frac{6}{40}\)

\(\Rightarrow B=\frac{-1}{15}\)

22 tháng 10 2017

de qua

22 tháng 10 2018

\(a,A=\dfrac{101}{100}+\dfrac{102}{100}+\dfrac{103}{100}+...+\dfrac{199}{100}\)

\(A=\dfrac{101+102+103+...+109}{100}\)

Xét tử số : \(101+102+103+...+199\)

Có : \(\left(199-101\right):1+1=99\) (số hạng)

\(\Rightarrow\) Tử số bằng \(:\left(199+101\right).99:2=14850\)

\(\Rightarrow A=\dfrac{14850}{100}=\dfrac{297}{2}\)

\(b,B=\dfrac{10002}{10000}+\dfrac{10004}{10000}+\dfrac{10006}{10000}+...+\dfrac{12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

Xét tử số : \(10002+10004+10006+...+12014\)

Có : \(\left(12014-10002\right):2+1=1007\) (số hạng)

\(\Rightarrow\) Tử số bằng : \(\left(12014+10002\right).1007:2=11085056\)

\(\Rightarrow B=\dfrac{11085056}{10000}\)

Bạn tự làm câu C nha

\(D=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{2015}=\dfrac{402}{2015}\)

\(E=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2014.2017}\)

\(\Rightarrow3E=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\)

\(\Rightarrow3E=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\)

\(\Rightarrow3E=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(\Rightarrow E=\dfrac{2016}{2017}:3=\dfrac{672}{2017}\)

22 tháng 10 2018

D = \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) +...+ \(\dfrac{1}{2014.2015}\)
D = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
+...+ \(\dfrac{1}{2014}-\dfrac{1}{2015}\)
D = \(\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
D = \(\dfrac{403}{2015}-\dfrac{1}{2015}\)
D = \(\dfrac{402}{2015}\)

15 tháng 8 2016
  • \(A=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)

          A= 1/3 + 1/4-1/4+1/5-1/5+1/6-1/6+1/7-1/7+1/8-1/8+1/9-1/9+1/10

          A=1/3+1/10

          A=13/30

15 tháng 8 2016

a,\(A=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{8.9}-\frac{1}{9.10}\)

      \(=\frac{1}{12}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

      \(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

       \(=\frac{1}{12}-\frac{1}{4}+\frac{1}{10}=\frac{5}{60}-\frac{15}{60}+\frac{6}{60}=\frac{-1}{15}\) 

Vậy \(A=\frac{-1}{15}\)