tính P=-1+1/2.1+1/3.2+1/4.3+..........+1/2018.2017+1/2018
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26 tháng 9 2018
Mình làm bừa nha đúng đúng sai sai bạn đừng giận nhé !
\(\frac{1}{2019.2018}-\frac{1}{2018.2017}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{2019}-\frac{1}{2018}...-\frac{1}{3}-\frac{1}{2}-\frac{1}{2}-\frac{1}{1}\)
\(=\frac{1}{2019}-\left(\frac{1}{2018}-\frac{1}{2018}\right)-..-\frac{1}{1}\)
\(=\frac{1}{2019}-0-\frac{1}{1}=\frac{1}{2019}-\frac{1}{1}\)
\(=-\frac{2018}{2019}\)
26 tháng 9 2018
\(\frac{1}{2019.2018}-\frac{1}{2018.2017}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=-\left(1-\frac{1}{2019}\right)=-\frac{2018}{2019}\)
27 tháng 1 2017
\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{1998.1999}\right)\)
\(=\frac{1}{2000.1999}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(=\frac{1}{2000.1999}-\left(1-\frac{1}{1999}\right)\)
\(=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow P+\frac{1997}{1999}=\frac{1}{1999.2000}-\frac{1998}{1999}+\frac{1997}{1999}\)
\(=\frac{-1}{2000}\)
27 tháng 1 2017
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 + \(\frac{1997}{1999}\)= \(\frac{-1997}{1999}-\frac{1}{2000}+\frac{1997}{1999}=\frac{-1}{2000}\)
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
\(a,1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=1-\left(1-\frac{1}{2014}\right)\)
\(=1-1+\frac{1}{2014}\)
\(=\frac{1}{2014}\)
24 tháng 11 2017
\(P=-1+\dfrac{1}{2.1}+\dfrac{1}{3.2}+..........+\dfrac{1}{2017.2016}+\dfrac{1}{2017}\)
\(=-1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{2016.2017}+\dfrac{1}{2017}\)
\(=-1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.........+\dfrac{1}{2016}-\dfrac{1}{2017}+\dfrac{1}{2017}\)
\(=-1+1-\dfrac{1}{2017}+\dfrac{1}{2017}\)
\(=0\)
3 tháng 1 2017
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)
3 tháng 1 2017
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+..+\frac{1}{3.2}+\frac{1}{2.1}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{1999}-\frac{1}{2000}\)
=\(1-\frac{1}{2000}\)
=\(\frac{1999}{2000}\)
TL
6
12 tháng 4 2019
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\right)\)
\(=\frac{1}{1.2}-\left(\frac{1}{2}-\frac{1}{99}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{99}\)
\(=\frac{1}{99}\)
NT
5
HT
17 tháng 7 2015
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{2.1}\)
=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)\)
=\(-\left(1-\frac{1}{100}\right)\)
=\(\frac{-99}{100}\)
17 tháng 7 2015
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
PT
0
\(P=\)\(-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)
\(P=-1+1-\frac{1}{2018}+\frac{1}{2018}\)
\(P=0\)
\(P=-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)
\(P=-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)
P = 0