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\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-1+\frac{1}{100}\)
\(C=\frac{-49}{50}\)
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
1) Tính:
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)\(-\frac{1}{2.1}\)
\(=\frac{1}{100}-\frac{1}{99}-\frac{1}{99}-\frac{1}{98}-\frac{1}{98}-\frac{1}{97}-\)\(...-\frac{1}{3}-\frac{1}{2}-\frac{1}{2}-\frac{1}{1}\)
\(=\frac{1}{100}-\frac{1}{1}\)
\(=-\frac{99}{100}\)
\(C=\frac{1}{100} -\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{2.1}\right)=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)
= (1/99-1/100)- (1/98-1/99)-...(1/1-1/2)
= -(1/1-1/2+1/3-1/4+...+1/99-1/100)
=-(1/1-1/100)
=-99/100
trong câu hỏi tương tự rõ hơn
\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+..+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
Đặt biểu thức trong ngoặc đơn là B ta có
\(B=\frac{100-99}{100.99}+\frac{99-98}{99.98}+\frac{98-97}{98.97}+...+\frac{3-2}{3.2}+\frac{2-1}{2.1}\)
\(B=\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+\frac{1}{2}-\frac{1}{3}+1-\frac{1}{2}\)
\(B=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}\)
#)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 = \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-.....-\frac{1}{2.1}\)
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 = \(\frac{-49}{50}\)
Ta có:\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{9900}-\left(\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100.99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)