\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{97.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\) 

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(\frac{1}{100}-C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)

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

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

\(C=C=\frac{1}{50}-1=-\frac{49}{50}\)

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!

\(C=\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)

\(C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\right)\)

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

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

\(C=\frac{1}{100}-\frac{99}{100}\)

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

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

\(=\frac{1}{100}-\left(\frac{1}{99.100}+\frac{1}{98.99}+\frac{1}{97.98}+...+\frac{1}{1.2}\right)\)

\(=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1-\frac{1}{2}\right)\)

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

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

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

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

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

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

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=\frac{-98}{100}=\frac{-49}{50}\)

        Bài làm :

Ta có :

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

\(C=\frac{1}{100}-\left(\frac{1}{99.100}+\frac{1}{98.99}+\frac{1}{97.98}+...+\frac{1}{1.2}\right)\)

\(C=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1-\frac{1}{2}\right)\)

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

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

Vậy C=-49/50

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

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

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

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

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

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

\(\)

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

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

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

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

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=-\frac{98}{100}=-\frac{49}{50}\)

Ủng hộ mk nha ^-^

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

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

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

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

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=\frac{-98}{100}=\frac{-49}{50}\)

Ủng hộ mk nha ^_-

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

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

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

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

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

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

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

\(c=\frac{1}{100}-\frac{1}{100.98}\frac{1}{99.98}\frac{1}{98.97}-......-\frac{1}{3.2}-\frac{1}{2.1}\)=\(\frac{1}{100}-\left[\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right]\) =\(\frac{1}{100}-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\right]\)=\(\frac{1}{100}-\left[1-\frac{1}{100}\right]=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{49}{50}\)