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A = 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/1.2
A = - (1/99.100 + 1/98.99 + 1/97.98 +... + 1/1.2)
A = - ( 1/99 - 1/100 + 1/98 - 1/99 + 1/97 - 1/98 +... + 1 - 1/2)
A = -(1 - 1/100)
A = -99/100
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
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
\(C=\frac{1}{100} -\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{2.1}\right)=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)
\(C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+\cdot\cdot\cdot+\frac{1}{2\cdot1}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(\frac{100}{100}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}\)
\(C=\frac{-98}{100}\)
\(Khi\) đó \(50C=\frac{-98}{100}\cdot50\)
\(50C=-49\)
tick cho mình nha
\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-\frac{1}{1}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{99}-1\right)=\frac{1}{99}-\frac{1}{99}+1=1\)
Ta có :
\(\frac{2015.2000-15}{2016.1999+1}\)
= \(\frac{2015.1999+2015-15}{2015.1999+1999+1}\)
= \(\frac{2015.1999+2000}{2015.1999+2000}\)
= 1
Vậy \(\frac{2015.2000-15}{2016.1999+1}=1\)
a)\(\frac{1}{2}-2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\)
=\(\frac{1}{50}\)
\(1)a)\frac{1}{2}-2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{24.25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{24}-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\frac{24}{25}=\frac{-23}{50}\)
\(\)
không đâu dễ mà
\(D=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}=\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)
đề có thiếu ko