1/100. 99- 1/99. 98- 1/98. 97.... 1/3.2-1/2.1 = ?
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LL
1
5 tháng 9 2018
đề :
= 1/100 - (1 / 100.99 +1/99.98 + ...+ 1/3.2 +1/2.1 )
=1/100 - (1 /1.2 +1/ 2.3 +...+ 1/ 98.99 +1 / 99.100)
=1/100 -( 1- 1/ 2 +1/2 -1/3 +...+1/98 -1/99 +1/99 -1/100)
=1/100 - ( 1- 1/100)
=1/100 - 99 /100
= -98/100
= -49 /50
TT
0
XO
13 tháng 9 2020
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2020.2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
b) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{21.23}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{21.23}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{21}-\frac{1}{23}\right)=\frac{1}{2}\left(1-\frac{1}{23}\right)=\frac{1}{2}.\frac{22}{23}=\frac{11}{23}\)
c) \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}=\frac{1}{99}-\left(\frac{1}{98.99}+\frac{1}{97.98}+...+\frac{1}{1.2}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1-\frac{1}{2}\right)=\frac{1}{99}-\left(-\frac{1}{99}+1\right)=\frac{1}{99}-\frac{98}{99}\)
\(=-\frac{97}{99}\)
d) bạn xem lại đề
13 tháng 9 2020
a)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=\frac{1}{1}-\frac{1}{2021}\)
\(=\frac{2020}{2021}\)
b)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{21\cdot23}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{23}\right)\)
\(=\frac{1}{2}\cdot\frac{22}{23}\)
\(=\frac{11}{23}\)
c)
\(=\frac{1}{99}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}\)
\(=\frac{-97}{99}\)
d)
đề sai hay sao á mong bạn xem ljai ạ
BJ
2
21 tháng 8 2016
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
21 tháng 8 2016
\(\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\frac{99}{100}\)
\(=\frac{-49}{50}\)
19 tháng 11 2017
Gọi \(A=\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow A=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(\Rightarrow A=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)
\(\Rightarrow A=\frac{1}{9900}-\frac{98}{99}=\frac{1}{9900}-\frac{9800}{9900}\)
\(\Rightarrow A=\frac{-9799}{9900}\)
TH
19 tháng 11 2017
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}=-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)
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
17 tháng 6 2016
1. 1-2+3-4+5-6-.....+99-100
=(1-2)+(3-4)+(5-6)+...+(99-100) (50 cặp)
=(-1)+(-1)+(-1)+...+(-1) (50 số -1)
=(-1).50
=-50
2.1+3-5-7+9+11-.....-397-399
=(1+3-5-7)+(9+11-13-15)+....+(387+389-391-393)+395-397-399 (99 cặp)
=(-8)+(-8)+(-8)+...+(-8)+(-401)(có 99 có -8)
=(-8).99+(-401)
=(-792)+(-401)
=-1193
3. 1-2-3+4+5-6-7+...+96+97-98-99+100
=(1-2-3+4)+(5-6-7+8)+...+(93-94-95+96)+(97-98-99+100) (25 cặp)
=0+0+0+...+0
=0
4. A=2100-299-298-.....-22-2-1
2A=2101-2100-299-....-23-22-2
2A-A=A=2101-2100-2100+1
A=2101-2.2100+1
A=2101-2101+1
A=1
NT
1
19 tháng 9 2018
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+\frac{1}{98\cdot97}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(\Rightarrow C=\frac{1}{100}-1+\frac{1}{100}\)
\(\Rightarrow C=\left(\frac{1}{100}+\frac{1}{100}\right)-1\)
\(\Rightarrow C=\frac{1}{50}-1\)
\(\Rightarrow C=\frac{-49}{50}\)
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