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 

\(C=\frac{1}{100} -\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{2.1}\right)=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)

chắc là 200,đoán thế thôi,chưa tính

​

\(= \frac{1}{99} - \left(\right. \frac{1}{99.98} + \frac{1}{98.97} + \frac{1}{97.96} + . . . + \frac{1}{3.2} + \frac{1}{2.1} \left.\right)\)

\(= \frac{1}{99} - \left(\right. \frac{1}{99} - \frac{1}{98} + \frac{1}{98} - \frac{1}{97} + \frac{1}{97} - \frac{1}{96} + . . . + \frac{1}{3} - \frac{1}{2} + \frac{1}{2} - \frac{1}{1} \left.\right)\)

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

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

\(C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)

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

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

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

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

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

\(Khi\) đó \(50C=\frac{-98}{100}\cdot50\)

\(50C=-49\)

tick cho mình nha

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-\frac{1}{1}\right)\)

\(=\frac{1}{99}-\left(\frac{1}{99}-1\right)=\frac{1}{99}-\frac{1}{99}+1=1\)

=1 nha bn, chắc vậy

=1

Đặt \(A=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-\cdots-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(A=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-\left(\frac{1}{96}-\frac{1}{97}\right)-\cdots-\left(\frac12-\frac13\right)-\left(\frac11-\frac12\right)\)

\(A=\frac{1}{99}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-\frac{1}{96}+\frac{1}{97}-\cdots-\frac12+\frac13-1+\frac12\)

\(A=\left(\frac{1}{99}-\frac{1}{99}\right)+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{97}-\frac{1}{97}\right)+\left(\frac{1}{96}-\frac{1}{96}\right)+\cdots+\left(\frac12-\frac12\right)-1\)

\(A=0+0+0+0+\cdots+0+\left(-1\right)\)

\(A=-1\)

Vậy A = -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 \(\frac{2015.2000-15}{2016.1999+1}=1\)