Lời giải:

Gọi tổng trên là $A$
$A=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)$

$=2\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{27-25}{25.27}\right)$

$=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{25}-\frac{1}{27}\right)$

$=2\left(1-\frac{1}{27})=\frac{52}{27}$

A = \(\dfrac{4}{1\times3}\) - \(\dfrac{8}{3\times5}\) + \(\dfrac{12}{5\times7}\) - \(\dfrac{16}{7\times9}\) + \(\dfrac{20}{9\times11}\) - \(\dfrac{24}{11\times13}\)

A = ( \(\dfrac{1}{1}+\dfrac{1}{3}\)) - ( \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\)+ \(\dfrac{1}{7}\)) - ( \(\dfrac{1}{7}\) + \(\dfrac{1}{9}\)) +( \(\dfrac{1}{9}\)+ \(\dfrac{1}{11}\)) - (\(\dfrac{1}{11}\)+\(\dfrac{1}{13}\))

A = \(\dfrac{1}{1}+\dfrac{1}{3}\) - \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}\) - \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{11}\) - \(\dfrac{1}{13}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{13}\)

A = \(\dfrac{12}{13}\)

\(=\frac{6}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-............+\frac{1}{97}-\frac{1}{99}\right).\\ \)

\(=\frac{6}{2}\left(1-\frac{1}{97}\right)\)

tới đây tính máy là ra luôn

sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

\(2\cdot\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right)\)

Theo quy luật :\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

                         \(2.\left(1-\frac{1}{15}\right)\)

                         \(2.\frac{14}{15}\)

                          \(\frac{28}{15}\)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{13\times15}+\dfrac{2}{15\times17}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=1-\dfrac{1}{17}\)

\(=\dfrac{16}{17}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{15\cdot17}\)

\(=2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=2-\dfrac{1}{17}\)

\(=\dfrac{33}{17}\)

Đặt Tổng trên là A

A     = 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/2005.2007

2. A = 2 . ( 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/2005.2007 )

2A   =  2/1.3  +  2/3.5  +  2/5.7  + ..... + 2/2005.2007

2A   =  1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/2005 - 1/2007

2A   =   1  -  1/2007

2A   =    2006/2007

 A     =  2006/2007 : 2

A      =  2006/4014

- Hok Tot - 

               \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+....+\dfrac{1}{2005\times2007}\)

=        \(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2005}-\dfrac{1}{2007}\right)\)

=        \(\dfrac{1}{2}\times\left(\dfrac{1}{1}-\dfrac{1}{2007}\right)\) 

=        \(\dfrac{1}{2}\times\dfrac{2006}{2007}\)

=              \(\dfrac{1003}{2007}\)

lớp 4 đã học cái này rồi hả

`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`

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)

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

\(=1-\frac{1}{9}=\frac{8}{9}\)

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)

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

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

\(=1-\frac{1}{9}=\frac{8}{9}\)

 ~ Hok tốt ~

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