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\(S=1+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+...+\dfrac{649}{650}\)

\(=1+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+...+1-\dfrac{1}{650}\)

\(=\left(1+1+...+1\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{650}\right)\)

\(=\left(1+1+...+1\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{25\cdot26}\right)\)

\(=25-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{25}-\dfrac{1}{26}\right)\)

\(=25-\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=25-\dfrac{12}{26}=25-\dfrac{6}{13}=\dfrac{319}{13}\)

4 tháng 4 2018

(đề bài)

=1-1/2+1-1/6+1-1/12+...+1-1/650

=(1+1+1+...+1)-(1/2+1/6+1/12+...+1/650)

=25-(1/1.2+1/2.3+1/3.4+...+1/25.26)

25-(1-1/2+1/2-1/3+1/3-1/4+...+1/25-1/26)

=25-(1-1/26)

=25/1-25/26

=625/26

    ĐS:625/26

 Hi!Hi!Kết bạn nhé!!!!!!!!!!

31 tháng 3 2019

tại sao:(1+1+1+...+1)=25

\(S=\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{600}+\dfrac{1}{650}\)

\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}\)

\(=\dfrac{1}{2}-\dfrac{1}{26}=\dfrac{13-1}{26}=\dfrac{12}{26}=\dfrac{6}{13}\)

7 tháng 6 2019

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

\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)

\(\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)8 số hạng 1

\(\left(1.8\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)

\(8-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(8-\left(1-\frac{1}{10}\right)\)

\(8-\frac{9}{10}\)

\(\frac{71}{10}\)

22 tháng 7 2017

Ta có: S = \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}\)

Áp dụng tính chất phép cộng.

Ta lại có S = \(\left(\frac{5}{6}+\frac{11}{12}\right)+\frac{19}{20}\)

\(\frac{21}{12}+\frac{19}{20}\)

\(\frac{648}{240}\)

22 tháng 7 2017

S=5/6+11/12+19/20

S=27/10

15 tháng 10 2018

\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{38}{5}\)

6 tháng 6 2021

100/11

hok tốt

6 tháng 8 2016

A = 1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72

A = ( 1 - 1/2 ) + ( 1 - 1/6 ) + ( 1 - 1/12 ) + ( 1 - 1/20 ) + ( 1 - 1/30 ) + ( 1 - 1/42 ) + ( 1 - 1/56 ) + ( 1 - 1/72 )

A = 1 x 8 - ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 )

A = 8 - ( \(\frac{1}{1\cdot2} +\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\))

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

A = \(8-\left(1-\frac{1}{9}\right)\)

\(A=8-\frac{8}{9}\)

\(A=\frac{64}{9}\)

12 tháng 5 2018

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15 tháng 10 2018

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

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

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

12 tháng 6 2016

Xem ở đây bạn nhé! http://olm.vn/hoi-dap/question/602318.html

S= 81/10.

1/2 + 5/6 + 11/12 + ... + 71/72 + 89/90 = 9 - ( 1/1x2 + 1/2x3 + ... + 1/9x10 ) = 9 - ( 1/1 - 1/2 + 1/2 - ... + 1/9 - 1/10 = 9 - ( 1 - 1/10 ) = 8 + 1/10 = 81/10