Tính : \(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{2.1}=?\)
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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 :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
LA
21 tháng 6 2017
1/100‐1/100.99‐1/99.98‐...‐1/3.2‐1/2.1
\(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=-\frac{49}{50}\)
HT
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
LK
6 tháng 7 2019
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}=...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right).\)
\(=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)
chúc bạn học tốt
8 tháng 9 2016
Ta có:\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{9900}-\left(\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)
HH
30 tháng 12 2017
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100.99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)
LN
1
NM
25 tháng 6 2016
Ta có: \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}\right)-\left(\frac{1}{98}-\frac{1}{99}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)
\(\Rightarrow C=\frac{1}{100}-\frac{1}{99}+\frac{1}{100}-\frac{1}{98}+\frac{1}{99}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(\Rightarrow C=\frac{1}{100}+\frac{1}{100}-1\)
\(\Rightarrow C=\frac{2}{100}-\frac{100}{100}\)
\(\Rightarrow C=-\frac{88}{100}=-\frac{22}{25}\)
Vậy \(C=-\frac{22}{25}\)
Chuk bạn hok tốt! 
29 tháng 6 2016
1) Tính:
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)\(-\frac{1}{2.1}\)
\(=\frac{1}{100}-\frac{1}{99}-\frac{1}{99}-\frac{1}{98}-\frac{1}{98}-\frac{1}{97}-\)\(...-\frac{1}{3}-\frac{1}{2}-\frac{1}{2}-\frac{1}{1}\)
\(=\frac{1}{100}-\frac{1}{1}\)
\(=-\frac{99}{100}\)
DG
4
2 tháng 9 2020
Các bạn giúp mk với, mk sắp phải nộp rồi. Ai nhanh nhất mk k cho
NN
2 tháng 9 2020
\(-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..........-\frac{1}{2.1}\)
\(=-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+..........+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{2.1}\)
\(\frac{1}{100-99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+..+\frac{1}{2.1}\right)\)
\(\frac{1}{100-99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(\frac{1}{100.99}-\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(\frac{1}{100.99}-\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\frac{1}{99}-\frac{1}{100}-\frac{98}{99}\)
\(-\frac{97}{99}-\frac{1}{100}\)
\(-\frac{9799}{9900}\)
\(\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{2\cdot1}\)
\(=\frac{1}{100\cdot99}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}\right)\)
\(=\frac{1}{99\cdot100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}\)
\(=\frac{-9799}{9900}\)