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19 tháng 11 2017

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

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

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

\(=\dfrac{1}{99.100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{99.100}-\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{99}-\dfrac{1}{100}-1-\dfrac{1}{99}\)

\(=-\dfrac{1}{100}-1=-\dfrac{101}{100}\)

20 tháng 11 2017

\(\Rightarrow=\dfrac{1}{100.99}-\left(\dfrac{1}{99.98}+\dfrac{1}{99.97}+...+\dfrac{1}{2.1}\right)\)

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

\(\Rightarrow\dfrac{1}{100}-\left(\dfrac{1}{99}-1\right)\)

\(\Rightarrow\dfrac{1}{100}-\dfrac{-98}{99}\)

=......... bn tính nhé

19 tháng 11 2017

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

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

\(\Rightarrow A=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)

\(\Rightarrow A=\frac{1}{9900}-\frac{98}{99}=\frac{1}{9900}-\frac{9800}{9900}\)

\(\Rightarrow A=\frac{-9799}{9900}\)

19 tháng 11 2017

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}=-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)

16 tháng 7 2015

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)

=\(\frac{1}{99}-\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}\right)\)

=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)

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

=\(\frac{1}{99}-\frac{97}{198}\)

=\(\frac{-95}{198}\)

30 tháng 10 2021

\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{2.1}\)

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

\(=\dfrac{2}{99}-\dfrac{1}{100}-1=-\dfrac{9799}{9900}\)

30 tháng 10 2021

Em chả hiểu j

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 :

\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)

26 tháng 7 2017

\(=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\) 

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

\(=\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)

28 tháng 8 2018

Ta có \(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)...... , \(\frac{1}{97.98}-\frac{1}{98.99}=\frac{2}{97.98.99}\)

vậy 2 xA = \(\frac{2}{1.2.3.}\) -\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)-.\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\).....-\(\frac{1}{97.98}\)+\(\frac{1}{98.99}\)

=1/3-1/6+1/(98.99) =1/6 +1/(98.99)

=> A = 1/12+\(\frac{1}{2.98.99}\)

21 tháng 6 2017

bài này dễ mak bn !tự lm đê!

21 tháng 6 2017

 1/100‐1/100.99‐1/99.98‐...‐1/3.2‐1/2.1

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

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

\(\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=-\frac{49}{50}\)

13 tháng 7 2016

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

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

\(C=\frac{1}{100}-1+\frac{1}{100}\)

\(C=\frac{-49}{50}\)

13 tháng 7 2016

C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1

C = 1/100 - (1/100.99 + 1/99.98 + 1/98.97 + ... + 1/3.2 + 1/2.1)

C = 1/100 - (1/1.2 + 1/2.3 + ... + 1/98.99 + 1/99.100)

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

C = 1/100 - (1 - 1/100)

C = 1/100 - 99/100

C = -98/100 = -49/50

23 tháng 2 2020

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

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

\(\frac{1}{99}+1=\frac{100}{99}\)

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

\(=-\left(\frac{1}{99}+\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

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

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

\(=-\frac{98}{99}\)