F = \(\dfrac{-1}{2016.2015}-\dfrac{1}{2015.2014}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
12 tháng 9 2021
\(\dfrac{1}{2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}=\dfrac{1}{2014}-\left(\dfrac{1}{2013.2014}+\dfrac{1}{2012.2013}+....+\dfrac{1}{1.2}\right)=\dfrac{1}{2014}-\left(\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2012}-\dfrac{1}{2013}+...+1-\dfrac{1}{2}\right)=\dfrac{1}{2014}-\left(1-\dfrac{1}{2014}\right)=\dfrac{1}{2014}-\dfrac{2013}{2014}=-\dfrac{2012}{2014}=-\dfrac{1006}{1007}\)
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 @@
20 tháng 9 2021
\(A=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ =\dfrac{1}{100}+1=\dfrac{101}{100}\)
OY
20 tháng 9 2021
\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(A=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(A=\dfrac{1}{100}-\dfrac{99}{100}=\dfrac{-49}{50}\)
VD
26 tháng 5 2022
\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-...-\dfrac{1}{2.1}\\ =\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\\ =\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\\ =\dfrac{1}{100}-\dfrac{99}{100}\\ =\dfrac{-98}{100}\\ =-\dfrac{49}{100}\)
26 tháng 5 2022
\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=-\dfrac{49}{50}\)
2 tháng 12 2017
Ta có B= \(\frac{1}{2009.2010}-(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}+\frac{1}{2008.2009}) \)
=\(\frac{1}{2009.}-\frac{1}{2010} -(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008} +\frac{1}{2008}-\frac{1}{2009}) \)
=\(\frac{1}{2009}-\frac{1}{2010}-(1-\frac{1}{2009} )\)
=\(\frac{2}{2009}-1 -\frac{1}{2010} \)
H
26 tháng 7 2017
Giải:
\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{99}-\left(\dfrac{1}{99.98}+\dfrac{1}{98.97}+\dfrac{1}{97.96}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{99}-\left(\dfrac{1}{99}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{96}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\right)\)
\(=\dfrac{1}{99}-\left(\dfrac{1}{99}-1\right)\)
\(=\dfrac{1}{99}-\dfrac{-98}{99}\)
\(=\dfrac{1}{99}+\dfrac{98}{99}\)
\(=\dfrac{99}{99}=1\)
Chúc bạn học tốt!
26 tháng 7 2017
\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}+\dfrac{1}{2.1}\)
=\(\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{97}+\dfrac{1}{96}-\dfrac{1}{96}+...+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)
=\(0+1\)
=\(1\)
Bạn học tốt^^
AH
Akai Haruma
Giáo viên
7 tháng 9 2018
Lời giải:
Đặt \(A=\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow A+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}=\frac{1}{99.100}\)
\(\Leftrightarrow A+\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}=\frac{1}{99.100}\)
\(\Leftrightarrow A+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}=\frac{1}{99.100}\)
\(\Leftrightarrow A+1-\frac{1}{98}=\frac{1}{99.100}\Rightarrow A=\frac{1}{9900}-\frac{97}{98}\)
\(F=-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\)
\(\Rightarrow-F=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}\)\(\Rightarrow F=\dfrac{-2015}{2016}\)
Giải:
\(F=\dfrac{-1}{2016.2015}-\dfrac{1}{2015.2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(\Leftrightarrow F=-\left(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+\dfrac{1}{2013.2012}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}+\dfrac{1}{2013.2014}+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2016}\right)\)
\(\Leftrightarrow F=-\dfrac{2015}{2016}\)
Vậy ...