1/2014 - 1/2014.2013 - 1/2013.2012 - ... - 1/3.2 - 1/2.1
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DK
1
11 tháng 9 2021
\(\dfrac{1}{2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}=\dfrac{1}{2014}-\left(\dfrac{1}{2013.2014}+\dfrac{1}{2012.2013}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)=\dfrac{1}{2014}-\left(\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2012}-\dfrac{1}{2013}+...+\dfrac{1}{2}-\dfrac{1}{3}+1-\dfrac{1}{2}\right)=\dfrac{1}{2014}-\left(1-\dfrac{1}{2014}\right)=\dfrac{1}{2014}-\dfrac{2013}{2014}=-\dfrac{1006}{1007}\)
DS
1
1 tháng 7 2023
=1/2014-(1/1*2+1/2*3+...+1/2013*2014)
=1/2014-(1-1/2+1/2-1/3+...+1/2013-1/2014)
=1/2014-1+1/2014
=1/1007-1=-1006/1007
12 tháng 9 2021
\(\dfrac{1}{2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}=\dfrac{1}{2014}-\left(\dfrac{1}{2013.2014}+\dfrac{1}{2012.2013}+....+\dfrac{1}{1.2}\right)=\dfrac{1}{2014}-\left(\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2012}-\dfrac{1}{2013}+...+1-\dfrac{1}{2}\right)=\dfrac{1}{2014}-\left(1-\dfrac{1}{2014}\right)=\dfrac{1}{2014}-\dfrac{2013}{2014}=-\dfrac{2012}{2014}=-\dfrac{1006}{1007}\)
DV
1
29 tháng 6 2021
\(=\frac{1}{2014}-\frac{2014-2013}{2014.2013}-\frac{2013-2012}{2013.2012}-...-\frac{3-2}{3.2}-\frac{2-1}{2.1}\)
\(=\frac{1}{2014}-\left(\frac{2014}{2014.2013}-\frac{2013}{2014.2013}\right)-...-\left(\frac{3}{3.2}-\frac{2}{3.2}\right)-\left(\frac{2}{2.1}-\frac{1}{2.1}\right)\)
\(=\frac{1}{2014}+\left(\frac{1}{2014}-\frac{1}{2013}\right)+...+\left(\frac{1}{3}-\frac{1}{2}\right)+\left(\frac{1}{2}-1\right)\)
\(=\frac{1}{1007}-1\)
\(=\frac{-1006}{1007}\)
CT
14 tháng 8 2021
Câu tl của bn lộn ở bước thứ 3 đấy đảo ngược 1/2013 và 1/2014 lại
27 tháng 9 2019
\(\frac{1}{2014}-\frac{1}{2014.2013}-\frac{1}{2013.2012}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}+\frac{1}{2013.2014}\right)+\frac{1}{2014}\)
\(=\frac{1}{2014}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=\frac{1}{2014}-1+\frac{1}{2014}=\frac{1}{1007}-1=\frac{-1006}{1007}\)
....
DT
0
21 tháng 9 2017
Ta có : \(1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-......-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2013.2014}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=1-\left(1-\frac{1}{2014}\right)\)
\(=1-1+\frac{1}{2014}\)
\(=\frac{1}{2014}\)
21 tháng 9 2017
\(a,1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=1-\left(1-\frac{1}{2014}\right)\)
\(=1-1+\frac{1}{2014}\)
\(=\frac{1}{2014}\)
LH
2
TL
17 tháng 6 2016
=\(-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2015}+\frac{1}{2014}-...-\frac{1}{2}+1\)
=\(-\frac{1}{2016}+1=\frac{2015}{2016}\)
CC
17 tháng 6 2016
Ta có :\(\frac{-1}{2016.2015}-\frac{1}{2015.2014}-\frac{1}{2014.2013}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(-\left(\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
= \(-\left(\frac{1}{2016}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2013}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-\frac{1}{1}\right)\)
= \(-\left(\frac{1}{2016}-1\right)\)
= \(-\left(-\frac{2015}{2016}\right)\)
= \(-\frac{2015}{2016}\)
Mk làm kĩ lắm rồi. ko tích nữa mk cũng chịu bạn luôn @@
PT
0
12014−12014.2013−12013.2012−...−13.2−12.1.12014−12014.2013−12013.2012−...−13.2−12.1.
=−(11.2+12.3+...+12012.2013+12013.2014)+12014=−(11.2+12.3+...+12012.2013+12013.2014)+12014
=12014−(1−12+12−13+...+12013−12014)=12014−(1−12+12−13+...+12013−12014)
=12014−1+12014=11007−1=−10061007
\(=\frac{1}{2014}-\frac{2014-2013}{2014\cdot2013}-\frac{2013-2012}{2013\cdot2012}-...-\frac{3-2}{3\cdot2}-\frac{2-1}{2\cdot1}\)
\(=\frac{1}{2014}-\left(\frac{2014}{2014\cdot2013}-\frac{2013}{2014\cdot2013}\right)-...-\left(\frac{3}{3\cdot2}-\frac{2}{3\cdot2}\right)-\left(\frac{2}{2\cdot1}-\frac{1}{2\cdot1}\right)\)
\(=\frac{1}{2014}+\left(\frac{1}{2014}-\frac{1}{2013}\right)+...+\left(\frac{1}{3}-\frac{1}{2}\right)+\left(\frac{1}{2}-1\right)\)
\(=\frac{1}{1007}-1\)
\(=-\frac{1006}{1007}\)