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22 tháng 3 2016

Ta có:

\(D=\frac{3^2}{1\times4}+\frac{3^2}{4\times7}+...+\frac{3^2}{88\times91}+\frac{3^2}{91\times94}\)

     \(=3\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{88\times91}+\frac{3}{91\times94}\right)\)

     \(=3\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{88}-\frac{1}{91}+\frac{1}{91}-\frac{1}{94}\right)\)

     \(=3\times\left(1-\frac{1}{94}\right)\)

     \(=3\times\frac{93}{94}\)

     \(=\frac{3\times93}{94}\)

     \(=\frac{279}{94}\)

Vậy \(D=\frac{279}{94}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

3 tháng 8 2016

\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+....+\frac{3}{25x28}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\)

\(1-\frac{1}{28}\)

\(\frac{27}{28}\)

k mình nha

18 tháng 10 2016

Bài 1 :

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)

Bài 2 :

\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)

\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)

Bài 3 :

\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)

\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)

\(3S=\frac{1}{4}-\frac{1}{22}\)

\(S=\frac{18}{88}\div3=\frac{6}{88}\)

23 tháng 4 2016

1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=1/2-1/3+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

=1/2-1/10

=5/10-1/10

=4/10=2/5

23 tháng 4 2016

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}-\frac{1}{10}\)

\(\frac{2}{5}\)

8 tháng 8 2017

\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)

\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)

8 tháng 8 2017

B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)

    = \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)

    = \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)

     = \(\frac{-22}{105}\)

C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)

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

     = \(1-\frac{1}{7}=\frac{6}{7}\)

19 tháng 7 2018

\(a,\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)

\(=\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)

\(=13\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{7.9}\right)\)

\(=13\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(=13.\frac{2}{9}=\frac{26}{9}\)

\(b,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}=\frac{2017}{2018}\)

P/s :Dấu chấm là dấu nhân nha

19 tháng 7 2018

phần c đâu bn