LQ
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TQ
2
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
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+..+\frac{1}{3.2}+\frac{1}{2.1}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{1999}-\frac{1}{2000}\)
=\(1-\frac{1}{2000}\)
=\(\frac{1999}{2000}\)
PT
0
27 tháng 8 2015
à mình nhầm có phải thế này không
1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1
=-(1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)
=-(1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )
=-(1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)
=-(1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)
=-(1/1-1/100)=-99/100
CT
1
30 tháng 9 2016
Đặt \(A=\frac{1}{2003.2004}-\frac{1}{2002.2003}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2002.2003}+\frac{1}{2003.2004}\)
\(\Rightarrow-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)
\(\Rightarrow-A=1-\frac{1}{2004}\)
\(\Rightarrow-A=\frac{2003}{2004}\)
\(\Rightarrow A=\frac{-2003}{2004}\)
26 tháng 6 2017
a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
\(A=1-\frac{1}{99}\)
\(A=\frac{98}{99}\)
thay A vào, ta được :
\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)
\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)
đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(A=2.\left(1-\frac{1}{99}\right)\)
\(A=2.\frac{98}{99}\)
\(A=\frac{196}{99}\)
Thay A vào, ta được :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
\(P=\)\(-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)
\(P=-1+1-\frac{1}{2018}+\frac{1}{2018}\)
\(P=0\)
\(P=-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)
\(P=-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)
P = 0