Bạn xem bài tương tự tại đây. Đề là:
Tính $(1+\frac{1}{1.3})(1+\frac{1}{2.4})....(1+\frac{1}{2021.2023})$

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)

\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(S=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(S=\frac{1}{2}.\frac{2016}{2017}\)

\(S=\frac{1008}{2017}< \frac{1}{2}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)

\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

\(2S=1-\frac{1}{2017}< 1\)

=> 2S < 1 

=> S < \(\frac{1}{2}\)(đpcm)

2016/2017 nhé 

k cho mình nha

cảm ơn bạn

\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{98.100}\)

\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}\)

\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)

\(S< 1-\frac{1}{99}< 1\)

=> S < 1

Cảm ơn bạn nhé

A= 2/1x3 + 2/3x5 + 2/5x7 +... + 2/2003x2005

A= 1 - 1/3 +1/3 - 1/5  + 1/5 - 1/7 + ... + 1/2003 + 1/2005

A= 1 - 1/2005

A= 2004/2005

B= 2006/2005

suy ra A < B

Ta co :A=2004/2005 vay thi A<B roi

\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{98}{99}=\dfrac{49}{99}>\dfrac{49}{100}=A\)