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DN
3
NQ
3 tháng 8 2015
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2006}-\frac{1}{2007}\)
\(C=1+0+0+...+0-\frac{1}{2007}\)
\(C=1-\frac{1}{2007}\)
\(C=\frac{2006}{2007}\)
LT
3
2 tháng 4 2019
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}+\frac{1}{2017\cdot2018}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2018}\)
\(=2-\frac{1}{2018}\)
\(=\frac{1009}{2018}-\frac{1}{2018}\)
\(=\frac{1008}{2018}=\)TỰ RÚT GỌN NHA
S
2 tháng 4 2019
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}+\frac{1}{2007.2008}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(=2-\frac{2007}{2008}\)
\(=\frac{2009}{2008}\)
~Học tốt~
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_
HA
2
19 tháng 7 2015
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2006\cdot2007}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}=1-\frac{1}{2007}=\frac{2006}{2007}\)
HT
19 tháng 7 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2006}-\frac{1}{2007}\)
=\(1-\frac{1}{2007}\)
=\(\frac{2006}{2007}\)
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!
6 tháng 5 2020
Đặt \(A=\frac{2006}{1\cdot2}+\frac{2006}{2\cdot3}+...+\frac{2006}{2006\cdot2007}\)
\(2006A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2006\cdot2007}\)
\(2006A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\)
\(2006A=\frac{1}{1}-\frac{1}{2007}\)
\(2006A=\frac{2006}{2007}\)
\(A=\frac{2006}{2007}\div2006\)
\(A=\frac{1}{2007}\)
Vậy giá trị của biểu thức bằng 1/2007
* Không chắc nha *
6 tháng 5 2020
Sửa đề : \(A=\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)
\(2006A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\)
\(2006A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\)
\(2006A=1-\frac{1}{2007}\)
\(2006A=\frac{2006}{2007}\)
\(A=\frac{2006}{2007}:2006=\frac{2006}{2007}.\frac{1}{2006}=\frac{1}{2007}\)
GL
2
2 tháng 2 2023
\(N=\dfrac{2006}{1.2}+\dfrac{2006}{2.3}+...+\dfrac{2006}{2006.2007}\)
\(N.2006=\dfrac{2006}{1}-\dfrac{2006}{2}+\dfrac{2006}{2}-\dfrac{2006}{3}+...+\dfrac{2006}{2006}-\dfrac{2006}{2007}\)
\(N.2006=2006-\dfrac{2006}{2007}\)
\(N=2006-\dfrac{2006}{2007}:2006\)
\(N=2006-\dfrac{1}{2007}\)
HT
3
18 tháng 3 2017
\(\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)
\(=2006.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\right)\)
\(=2006.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\right)\)
\(=2006.\left(1-\frac{1}{2007}\right)\)
\(=2006.\frac{2006}{2007}\)
\(=\frac{2006^2}{2007}\)
HL
18 tháng 3 2017
\(=\frac{2006}{1.2}+\frac{2006}{2.3}+...+\frac{2006}{2006.2007}\)
\(=2006 \left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\right)\)
\(=2006.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\right)\)
\(=2006.\left(1-\frac{1}{2007}\right)\)
\(=2006.\frac{2006}{2007}=\frac{4024036}{2007}\)
NH
10 tháng 3 2023
a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{2006.2007}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{2006}-\dfrac{1}{2007}\)
\(=1-\dfrac{1}{2007}=\dfrac{2006}{2007}\)
la sao