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\(\frac{1}{2019.2018}-\frac{1}{2018.2017}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{2019}-\frac{1}{2018}...-\frac{1}{3}-\frac{1}{2}-\frac{1}{2}-\frac{1}{1}\)

\(=\frac{1}{2019}-\left(\frac{1}{2018}-\frac{1}{2018}\right)-..-\frac{1}{1}\)

\(=\frac{1}{2019}-0-\frac{1}{1}=\frac{1}{2019}-\frac{1}{1}\)

\(=-\frac{2018}{2019}\)

\(\frac{1}{2019.2018}-\frac{1}{2018.2017}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=-\left(1-\frac{1}{2019}\right)=-\frac{2018}{2019}\)

\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{1998.1999}\right)\)

\(=\frac{1}{2000.1999}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)

\(=\frac{1}{2000.1999}-\left(1-\frac{1}{1999}\right)\)

\(=\frac{1}{1999.2000}-\frac{1998}{1999}\)

\(\Rightarrow P+\frac{1997}{1999}=\frac{1}{1999.2000}-\frac{1998}{1999}+\frac{1997}{1999}\)

\(=\frac{-1}{2000}\)

P= \(\frac{1}{2000.1999}\)-  (\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\))

  = \(\frac{1}{1999}-\frac{1}{2000}\)- (\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\))

  = \(\frac{1}{1999}-\frac{1}{2000}\)- ( \(1-\frac{1}{1999}\))

  = \(\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)

  = \(\frac{-1997}{1999}-\frac{1}{2000}\)

 =) P + \(\frac{1997}{1999}\)= \(\frac{-1997}{1999}-\frac{1}{2000}+\frac{1997}{1999}=\frac{-1}{2000}\)

thằng kia copy bài mình đó đừng tk nó

Ta có : \(1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-......-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2013.2014}\right)\)

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

\(=1-\left(1-\frac{1}{2014}\right)\)

\(=1-1+\frac{1}{2014}\)

\(=\frac{1}{2014}\)

\(a,1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)

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

\(=1-\left(1-\frac{1}{2014}\right)\)

\(=1-1+\frac{1}{2014}\)

\(=\frac{1}{2014}\)

1/2014

\(P=-1+\dfrac{1}{2.1}+\dfrac{1}{3.2}+..........+\dfrac{1}{2017.2016}+\dfrac{1}{2017}\)

\(=-1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{2016.2017}+\dfrac{1}{2017}\)

\(=-1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.........+\dfrac{1}{2016}-\dfrac{1}{2017}+\dfrac{1}{2017}\)

\(=-1+1-\dfrac{1}{2017}+\dfrac{1}{2017}\)

\(=0\)

\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)

\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)

\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+..+\frac{1}{3.2}+\frac{1}{2.1}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{1999}-\frac{1}{2000}\)

=\(1-\frac{1}{2000}\)

=\(\frac{1999}{2000}\)

$P = \dfrac1{2000 \cdot 1999} + \dfrac1{1999 \cdot 1998} + \ldots + \dfrac1{3 \cdot 2} + \dfrac1{2 \cdot 1} \\
= \dfrac1{1999} - \dfrac1{2000} + \dfrac1{1998} - \dfrac1{1999} + \ldots + \dfrac12 - \dfrac13 + \dfrac11 - \dfrac12
= - \dfrac1{2000} + \dfrac11 \\
= \dfrac{1999}{2000}$

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

=1-1/99=98/99

\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\right)\)

\(=\frac{1}{1.2}-\left(\frac{1}{2}-\frac{1}{99}\right)\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{99}\)

\(=\frac{1}{99}\)

ê Hoàng sai rồi 

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{2.1}\)

=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)\)

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

=\(\frac{-99}{100}\)

A=1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1

A=   - (1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)

A=   - (1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )

A=   - (1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)

A=   - (1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)

A=   - (1/1-1/100)

A=   - 99/100

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