Tính nhanh:
A= 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 -1/2.1
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HT
1
MA
14 tháng 9 2016
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{97.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(\frac{1}{100}-C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)
\(\frac{1}{100}-C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{100}-C=1-\frac{1}{100}\)
\(C=C=\frac{1}{50}-1=-\frac{49}{50}\)
ND
1
7 tháng 9 2016
C=1/100-(1/100.99+1/99.98+...+1/3.2+1/2.1)
=1/100-(1-1/2+1/2_1/3+...+1/99-1/100)
=1/100-(1-1/100)
=1/100-99/100
=1/100 chọn cho mình nha!
ND
0
LV
0
ND
1
7 tháng 9 2016
\(C=\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}\)
\(C=\frac{-49}{50}\)
KL
3
XO
13 tháng 9 2020
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{99.100}+\frac{1}{98.99}+\frac{1}{97.98}+...+\frac{1}{1.2}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1-\frac{1}{2}\right)\)
\(=\frac{1}{100}-\left(-\frac{1}{100}+1\right)\)
\(=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=-\frac{49}{50}\)
13 tháng 9 2020
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}\)
\(C=\frac{-98}{100}=\frac{-49}{50}\)
NL
0
WJ
4
24 tháng 7 2015
C= \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
= \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\) ( viet nguoc lai cho de nhin)
= \(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
= \(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
= \(-\frac{49}{50}\)
24 tháng 7 2015
C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - .... - 1/3.2 - 1/2.1
\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{2.1}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+...+1-\frac{1}{2}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{2}\right)=-\frac{1}{2}\)
NG
1
10 tháng 1 2016
50 nếu ai thích sakura thì **** mình nếu ai thích sakura mà Ko **** mình thì chứng tỏ bạn Ko thích sakura
\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}+\frac{1}{100}-1=\frac{1}{50}-1=-\frac{49}{50}\)
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
C = - 49/50