A = 1/3 - 1/4 + 1/4 -1/5 + 1/5 ....-1/20-1/21 

A = 1/3 - 1/21

a,=1/3-1/4+1/4-1/5+.............+1/20-1/21

  =1/3-1/21

=2/7

b,=1/2(1/4.6+1/6.8+............+1/30.32)

    =1/2(1/4-1/6+1/5+1/8+.............+1/30-1/32)

    =1/2(1/4-1/32)

    =1/2.7/32

    =1/64

a) 1/3-1/4+1/4-1/5...........+1/19-1/20+1/20-1/21=1/3-1/21

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2011 - 1/2012

A = 1 - 1/2012

A = 2011/2012

B = 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 +...+ 1/2010 - 1/2012

B = 1/2 - 1/2012

B = 1005/2012

a) \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)

\(A=1-\dfrac{1}{2012}\)

\(A=\dfrac{2011}{2012}\)

b) \(B=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2010\cdot2012}\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2010\cdot2012}\right)\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2010}-\dfrac{1}{2012}\right)\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2012}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{1005}{2012}\)

\(B=\dfrac{1005}{4024}\)

a, 1/1.2+1/2.3+1/3.4+...+1/999.1000

=  1/1-1/2+1/2-1/3+1/3-1/4+....+1/999-1/1000

=   1/1-1/1000

=   999/1000

b, 1/2.4+1/4.6+1/6.8+1/8.10

=  1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10

=  1/2-1/10

=   4/10  =2/5

\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(2A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(2A=\frac{1}{1}-\frac{1}{100}\)

\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)

Câu B và C làm tương tự.

bạn Nhi làm sai rồi

\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được

\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)

kết quả là : \(\frac{49}{100}\)

\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

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

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)

A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2106}\)

\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)

\(A=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)

\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2014.2016}=\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1007.1008}\right)\)

=> \(B=\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{1008}\right)=\frac{1}{4}.\frac{1007}{1008}\)

=> \(B=\frac{1007}{4032}\)

\(A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(A=\frac{1}{2}.\frac{1009}{2020}\)

\(A=\frac{1009}{4040}\)

A=1/2.4+1/4.6+1/6.8+...+1/2018.2020

  =1/2(1/2-1/4+1/4-1/6+...+1/2018-1/2020)

    =1/2(1/2-1/2020)

   =1/2.1009/2020

   =1009/4040

#)Giải :

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2018.2020}\)

\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\)

     \(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)

     \(2A=\frac{1}{2}-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{6}-\frac{1}{6}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{2018}-\frac{1}{2018}\right)-\frac{1}{2020}\)

     \(2A=\frac{1}{2}-0-0-0-...-0-\frac{1}{2020}\)

     \(2A=\frac{1}{2}-\frac{1}{2020}\)

     \(2A=\frac{1009}{2020}\)

\(\Rightarrow A=\frac{1009}{4040}\)

#)Chúc bn học tốt :D