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3 tháng 8 2015

\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2006}-\frac{1}{2007}\)

\(C=1+0+0+...+0-\frac{1}{2007}\)

\(C=1-\frac{1}{2007}\)

\(C=\frac{2006}{2007}\)

17 tháng 2 2019

C=1-1/2+1/2-1/3+1/3-1/4+.......+1/2016-1/2017

gạch phân số giống  nhau

C=1-1/2017

C=2016/2017

1 tháng 12 2019

a, \(A=\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\)

  \(A=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\)

 \(A=\frac{1}{11}-\frac{1}{66}\)

\(A=\frac{5}{66}\)

b, \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

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

\(B=1-\frac{1}{7}\)

\(B=\frac{6}{7}\)

_Học tốt nha_

19 tháng 7 2015

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2006\cdot2007}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}=1-\frac{1}{2007}=\frac{2006}{2007}\)

19 tháng 7 2015

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2006}-\frac{1}{2007}\)

=\(1-\frac{1}{2007}\)

=\(\frac{2006}{2007}\)

16 tháng 3 2021

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{2006.2007}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{2006}-\dfrac{1}{2007}\)

\(=1-\dfrac{1}{2007}=\dfrac{2006}{2007}\)

16 tháng 3 2021

la sao

 

12 tháng 1 2017

Bài 2:

a) \(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)

\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}\)

\(=\frac{5}{66}\)

b) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

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

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

c) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\)

\(=1-\frac{1}{2007}\)

\(=\frac{2006}{2007}\)

15 tháng 1 2017

Bài 2:

a) \(\frac{5}{11.16}\) + \(\frac{5}{16.21}\) + \(\frac{5}{21.26}\) + ... + \(\frac{5}{61.66}\)

= \(\frac{1}{11}\) - \(\frac{1}{16}\) + \(\frac{1}{16}\) - \(\frac{1}{21}\) + \(\frac{1}{21}\) - \(\frac{1}{26}\) + ... + \(\frac{1}{61}\) - \(\frac{1}{66}\)

= \(\frac{1}{11}\) - \(\frac{1}{66}\)

= \(\frac{5}{66}\)

b) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

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

= \(1-\frac{1}{7}\)

= \(\frac{6}{7}\)

c) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1989.1990}+...+\frac{1}{2006.2007}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}+...+\frac{1}{2006}-\frac{1}{2007}\)

= \(1-\frac{1}{2007}\)

= \(\frac{2006}{2007}\)

Chúc bạn học tốt!

12 tháng 7 2016

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{200}-\frac{1}{201}\)

\(=1-\frac{1}{201}\)

\(=\frac{200}{201}\)

12 tháng 7 2016

 \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{200}-\frac{1}{201}\)

\(=1-\frac{1}{201}=\frac{200}{201}\)

Ủng hộ nha,tớ ko ăn cóp đâu.

7 tháng 3 2019

B=1/1-1/100 -1=-1/100

10 tháng 8 2018

100 số hạng đầu là

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}\)

ta có \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}\)\(=\frac{100}{101}\)

18 tháng 3 2017

\(\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)

\(=2006.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\right)\)

\(=2006.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\right)\)

\(=2006.\left(1-\frac{1}{2007}\right)\)

\(=2006.\frac{2006}{2007}\)

\(=\frac{2006^2}{2007}\)

18 tháng 3 2017

\(=\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)

\(=2006 \left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\right)\)

\(=2006.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\right)\)

\(=2006.\left(1-\frac{1}{2007}\right)\)

\(=2006.\frac{2006}{2007}=\frac{4024036}{2007}\)