A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

A = \(\frac{1}{2}.\left(1-\frac{1}{11}\right)=\frac{1}{2}.\frac{10}{11}\)

A = \(\frac{5}{11}\)

Ta có: \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2019\cdot2021}\right)\)

\(=\left(1+\frac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\frac{1}{\left(3-1\right)\left(3+1\right)}\right)\cdot\ldots\cdot\left(1+\frac{1}{\left(2020-1\right)\left(2020+1\right)}\right)\)

\(=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{2020^2-1}\right)\)

\(=\frac{2^2-1+1}{2^2-1}\cdot\frac{3^2-1+1}{3^2-1}\cdot\ldots\cdot\frac{2020^2-1+1}{2020^2-1}\)

\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{2020^2}{2019\cdot2021}\)

\(=\frac{2\cdot3\cdot\ldots\cdot2020}{1\cdot2\cdot\ldots\cdot2019}\cdot\frac{2\cdot3\cdot\ldots\cdot2020}{3\cdot4\cdot\ldots\cdot2021}=\frac{2020}{1}\cdot\frac{2}{2021}=\frac{4040}{2021}\)

 A = 1/1x3 + 1/3x5 + 1/5x7 +.........+ 1/2009x2011

    = 1/1-1 +1/3-5 + 1/5-7 + .......+ 1/2009-2011

    = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +........+ 1/2009 -1/2011

    = 1/1 - 1/2011

    = 2010/2011

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{100}{101}\)

\(=\frac{50}{101}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)

\(=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\right)\)

\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)

                            Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

                              \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

                             \(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

                            \(2A=1-\frac{1}{9.11}=1-\frac{1}{99}=\frac{98}{99}\)

                              \(A=\frac{98}{99}:2=\frac{49}{99}\)

                                Ủng hộ mk nha!!!

`1/[1xx3]+1/[3xx5]+1/[5xx7]+...+1/[17xx19]`

`=1/2xx(2/[1xx3]+2/[3xx5]+....+2/[17xx19])`

`=1/2xx(1-1/3+1/3-1/5+....+1/17-1/19)`

`=1/2xx(1-1/19)`

`=1/2xx18/19`

`=9/19`

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

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

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

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

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

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

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

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

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

=> 2S < 1 

=> S < \(\frac{1}{2}\)(đpcm)

\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)

=> \(\frac{5}{11}y=\frac{2}{3}\)

=>y = \(\frac{2}{3}:\frac{5}{11}\)

=> y = \(\frac{22}{15}\)

cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm

Giải:

\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)y=-\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}.\dfrac{10}{11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{5}{11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow15y=-22\)

\(\Leftrightarrow y=-\dfrac{22}{15}\)

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