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4 tháng 8 2016

\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)

\(A=1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)

\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=9-\left(1-\frac{1}{10}\right)=9-1+\frac{1}{10}=8\frac{1}{10}\)

23 tháng 2 2016

\(\frac{1}{2}+\frac{5}{6}+...+\frac{55}{56}\)

\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{56}\right)\)

\(=7.1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}\right)\)

\(=7-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\right)\)

\(=7-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(=7-\left(1-\frac{1}{8}\right)\)

\(=7-\frac{7}{8}=\frac{49}{8}\)

23 tháng 2 2016

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}=\frac{1889}{312}\)

6 tháng 3 2016

\(\frac{5}{14}\)

6 tháng 3 2016

A=\(\frac{1}{2}\)+\(\frac{5}{6}\)+\(\frac{11}{12}\)+\(\frac{19}{20}\)+\(\frac{29}{30}\)+\(\frac{41}{42}\)=(1-\(\frac{1}{2}\))+(1-\(\frac{1}{6}\))+(1-\(\frac{1}{12}\))+(1-\(\frac{1}{20}\))+(1-\(\frac{1}{30}\))+(1-\(\frac{1}{42}\))

  =1-1+\(\frac{1}{2}\)+1-\(\frac{1}{2}\)+\(\frac{1}{3}\)+1-\(\frac{1}{3}\)+\(\frac{1}{4}\)+1-\(\frac{1}{4}\)+\(\frac{1}{5}\)+1-\(\frac{1}{5}\)+\(\frac{1}{6}\)+1-\(\frac{1}{6}\)+\(\frac{1}{7}\)

  =(1-1+1+1+1+1+1)+(\(\frac{1}{2}\)-\(\frac{1}{2}\))+(\(\frac{1}{3}\)-\(\frac{1}{3}\))+(\(\frac{1}{4}\)-\(\frac{1}{4}\))+(\(\frac{1}{5}\)-\(\frac{1}{5}\))+(\(\frac{1}{6}\)-\(\frac{1}{6}\))+\(\frac{1}{7}\)

  =5+\(\frac{1}{7}\)=\(\frac{36}{7}\)

5 tháng 8 2015

\(A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

=>\(A=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

=>\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

=>\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

=>\(a=9-\left(1-\frac{1}{10}\right)=\frac{90}{10}-\frac{9}{10}=\frac{81}{10}\)

5 tháng 8 2015

9 - A = \(1-\frac{1}{2}+1-\frac{5}{6}+1-\frac{11}{12}+..+1-\frac{89}{90}=\frac{1}{2}+\frac{1}{6}+..+\frac{1}{90}\)

         = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

=> A = \(9-\frac{9}{10}=\frac{81}{10}\)

31 tháng 8 2015

=> A = ( 1 - 1/2 ) + ( 1 - 1/6 ) + ( 1 - 1/12 ) + ( 1 - 1/30 ) + .... + ( 1 - 1/90 )

=> A = ( 1 + 1 + 1 + .... 1 ) - ( 1/2 + 1/6 + 1/12 + 1/30 + .... + 1/90 )

=> A = 9 - ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/9.10 )

=> A = 9 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10 )

=> A = 9 - ( 1 - 1/10 )

=> A = 9 - 9/10

=> 81/10

17 tháng 2 2017

81/10

3 tháng 5 2016

A=1-1/2+1-1/6+...+1-1/90

  =9-(1/2+1/6+...+1/90) =9-(1/1.2+1/2.3+...+1/9.10)

  =9-(1-1/10)=9-9/10=81/10

26 tháng 7 2015

Ta có công thức : \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)

\(\Rightarrow B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)

           \(=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+10\right)10}{2}}\)

           \(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)

           \(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

           \(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

           \(=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)