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\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-....-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\frac{1}{99}+1=\frac{100}{99}\)
\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{99}+\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{99}-1\right)\)
\(=-\frac{98}{99}\)
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-1+\frac{1}{100}\)
\(C=\frac{-49}{50}\)
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
\(\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)
\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)
Đúq nhaaa
#)Giải :
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+... +\frac{1}{99.100}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=-\left(1-\frac{1}{100}\right)\)
\(=-\frac{99}{100}\)
#~Will~be~Pens~#
\(\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(=-\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right]\)
\(=-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right]\)
\(=-\left[1-\frac{1}{100}\right]=-\frac{99}{100}\)
\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+..+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
Đặt biểu thức trong ngoặc đơn là B ta có
\(B=\frac{100-99}{100.99}+\frac{99-98}{99.98}+\frac{98-97}{98.97}+...+\frac{3-2}{3.2}+\frac{2-1}{2.1}\)
\(B=\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+\frac{1}{2}-\frac{1}{3}+1-\frac{1}{2}\)
\(B=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}\)
C = \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-.....-\frac{1}{2.1}\)
C = \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
C = \(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
C = \(\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}\)
C = \(\frac{-49}{50}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)