\(\dfrac{1}{2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}=\dfrac{1}{2014}-\left(\dfrac{1}{2013.2014}+\dfrac{1}{2012.2013}+....+\dfrac{1}{1.2}\right)=\dfrac{1}{2014}-\left(\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2012}-\dfrac{1}{2013}+...+1-\dfrac{1}{2}\right)=\dfrac{1}{2014}-\left(1-\dfrac{1}{2014}\right)=\dfrac{1}{2014}-\dfrac{2013}{2014}=-\dfrac{2012}{2014}=-\dfrac{1006}{1007}\)

Giúp mình với 

\(\dfrac{1}{2014}-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-...-\dfrac{1}{2013\cdot2014}\)

\(=\dfrac{1}{2014}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\right)\)

\(=\dfrac{1}{2014}-1+\dfrac{1}{2014}=-\dfrac{1006}{1007}\)

=\(-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2015}+\frac{1}{2014}-...-\frac{1}{2}+1\)

=\(-\frac{1}{2016}+1=\frac{2015}{2016}\)

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 @@

=-1/99-(1-1/2+1/2-1/3+...+1/98-1/99)

=-2/99+1=97/99

\(A=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ =\dfrac{1}{100}+1=\dfrac{101}{100}\)

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(A=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(A=\dfrac{1}{100}-\dfrac{99}{100}=\dfrac{-49}{50}\)

A=1/100−1/100+1/99−1/99+1/98−1/98+1/97−...−1/3+1/2−1/2+1

=1

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-...-\dfrac{1}{2.1}\\ =\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\\ =\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\\ =\dfrac{1}{100}-\dfrac{99}{100}\\ =\dfrac{-98}{100}\\ =-\dfrac{49}{100}\)

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(=-\dfrac{49}{50}\)

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-...-\dfrac{1}{2.1}\\ =\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-...-\dfrac{1}{2}+1\\ =1\)

Ta có B= \(\frac{1}{2009.2010}-(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}+\frac{1}{2008.2009}) \)

=\(\frac{1}{2009.}-\frac{1}{2010} -(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008} +\frac{1}{2008}-\frac{1}{2009}) \)

=\(\frac{1}{2009}-\frac{1}{2010}-(1-\frac{1}{2009} )\)

=\(\frac{2}{2009}-1 -\frac{1}{2010} \)

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!

\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}+\dfrac{1}{2.1}\)

=\(\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{97}+\dfrac{1}{96}-\dfrac{1}{96}+...+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)

=\(0+1\)

=\(1\)

Bạn học tốt^^

Lời giải:

Đặt \(A=\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow A+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}=\frac{1}{99.100}\)

\(\Leftrightarrow A+\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}=\frac{1}{99.100}\)

\(\Leftrightarrow A+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}=\frac{1}{99.100}\)

\(\Leftrightarrow A+1-\frac{1}{98}=\frac{1}{99.100}\Rightarrow A=\frac{1}{9900}-\frac{97}{98}\)