Giải:

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2009}\)

\(=\dfrac{2008}{2009}\)

c) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{4}{7.10}+...+\dfrac{3}{94.97}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(=\dfrac{1}{1}-\dfrac{1}{97}\)

\(=\dfrac{96}{97}\)

Vậy ...

Các câu sau tương tự

b, \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2008.1009}\)

\(=\)\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2009}=\dfrac{2009}{2009}-\dfrac{1}{2009}=\dfrac{2008}{2009}\)

c,\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+....+\dfrac{3}{94.97}\)

\(=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+....+\dfrac{97-94}{94.97}\)

\(=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{94}-\dfrac{1}{97}\)

\(=\dfrac{1}{1}-\dfrac{1}{97}=\dfrac{97}{97}-\dfrac{1}{97}=\dfrac{96}{97}\)

\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...............+\dfrac{2}{2008.2009}\)

\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+................+\dfrac{1}{2008.2009}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.................+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)

\(=2\left(1-\dfrac{1}{2009}\right)\)

\(=2.\dfrac{2008}{2009}=\dfrac{4016}{2009}\)

Ta có : 1.2 + 2.3 + 3.4 + ... + 2008.2009

= ( 1.2.3 + 2.3.3 + 3.4.3 + ... + 2008.2009.3 ) :3

= [ 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ... + 2008.2009.( 2010 - 2007 )] : 3

= [ 1.2.3 + 2.3.4 - 2.3.1 + 2.4.5 - 3.4.2 + ... + 2008.2009.2010 - 2008.2009.2007 ] : 3

= ( 2008.2009.2010 ) :3

= 2702828240

1.2 + 2.3 + 3.4 + 4.5 +...+ 2008.2009 

= \(\frac{1}{3}\left(1.2.3+2.3.3+3.4.3+4.5.3+...+2008.2009.3\right)\)

= \(\frac{1}{3}\left(1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+2008.2009.\left(2010-2007\right)\right)\)

\(=\frac{1}{3}.2008.2009.2010=670.2008.2009\) số lớn nên bạn tự tính tiếp nhé!

Đặt A = 1.2 + 2.3 + 3.4 + ..... + 2008.2009

<=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2008.2009.3

<=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ...... + 2008.2009.( 2010 - 2007 )

<=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 2008.2009.2010 - 2007.2008.2009

<=> 3A = 2008.2009.2010

=> A = ( 2008.2009.2010 ) : 3

=132546879

kho

Đúng rồi đấy bạn

a=2009(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.........+1/2008-1/2009)

=2009x2008/2009

=2008

= 2/1 - 2/2 + 2/2 - 2/3 + 2/3 - 2/4 + ..... + 2/99 - 2/100

= 2/1 + 2/100

= 101/50

2/1 - 2/2 + 2/2 - 2/3 + 2/3 - 2/4 +...+ 2/99 - 2/100

= 2/1 - 2/100

=  99/50