\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{2.2004}{2010}=\frac{2004}{1005}\)

\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{1004\cdot1005}\)

\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1004\cdot1005}\right)\)

\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)

\(=2\cdot\left(1-\frac{1}{1005}\right)=2\cdot\frac{1004}{1005}=\frac{2008}{1005}\)

Gọi tổng là A ta có :

A x 2 = 2/2.4 + 2/4.6 + 2/6.8 + ... + 2/18.20

A x 2 = 1/2 - 1/4 - 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/18 - 1/20

A x 2 = 1/2 - 1/20

A x 2 = 9/20

      A = 9/20 : 2

      A = 9/40

9/40 nha 

9/40 nha 

Đặt:A =  \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)

=> A = 2.(\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{2008.2010}\)

=> A = 2.(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2008}-\frac{1}{2010}\)

=> A = 2.(\(\frac{1}{2}-\frac{1}{2010}\))

=> A = 2.\(\frac{502}{1005}\)

=> A = \(\frac{1004}{1005}\)

đặt A= \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}\) 

=> 1/2.A=\(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\) 

= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)

=\(\frac{1}{2}-\frac{1}{2010}\)

=\(\frac{502}{1005}\)

Vậy biểu thức cần tìm có giá trị là \(\frac{502}{1005}\)

nhầm nha

=> A= \(\frac{502}{1005}:\frac{1}{2}\)=\(\frac{1004}{1005}\)

\(B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{98.100}\)

\(\Rightarrow5B=\dfrac{20}{2.4}+\dfrac{20}{4.6}+...+\dfrac{20}{98.100}\)

\(\Rightarrow5B=10\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}\right)\)

\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(\Rightarrow5B=10.\dfrac{49}{100}\)

\(\Rightarrow5B=\dfrac{49}{10}\)

Vậy \(5B=\dfrac{49}{10}\)

Ta có: B = \(\dfrac{4}{2.4}\) + \(\dfrac{4}{4.6}\) + \(\dfrac{4}{6.8}\) + ... + \(\dfrac{4}{98.100}\).

=> \(\dfrac{B}{2}\) = \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + \(\dfrac{2}{6.8}\) + ... + \(\dfrac{2}{98.100}\)

=\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\)

= \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\) = \(\dfrac{49}{100}\).

=> B = \(\dfrac{49}{200}\).

=> 5B = \(\dfrac{49}{200}\) . 5 = \(\dfrac{49}{40}\).

Vậy 5B = \(\dfrac{49}{40}\).

Ta gọi biểu thức đó là A

Ta có công thức        \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)

Dựa vào công thức ta có  

\(\frac{4}{2.4}=2.\left(\frac{1}{2}-\frac{1}{4}\right)\)

\(\frac{4}{4.6}=2.\left(\frac{1}{4}-\frac{1}{6}\right)\)

\(....................\)

\(\frac{4}{18.20}=2.\left(\frac{1}{18}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{18}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{9}{20}\right)=\frac{18}{20}\)

Ai thấy đúng thì ủng hộ nha !!!

sai rồi kết quả phải bằng 9/10 chứ 

\(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{18\cdot20}\)

\(=2\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=2\left(\frac{9}{20}\right)\)\(=\frac{18}{20}=\frac{9}{10}\)

ax4=1/2-1/4+1/4-1/6+1/6+1/8+.......+1/16-1/18+1/18+1/20

ax4  =1/2-1/20

 ax4 =9/20

   a=9/20:4

   a=9/80

THANKS

K = đề bài

   = 2 . ( 2/2.4 + 2/4.6 + 2/6.8 + . . . + 2/2008.2010 )

   = 2 . ( 1 - 1/4 + 1/4 - 1/6 + 1/8 - 1/8 + . . . + 2/2008 - 2/2010 )

   = 2 . ( 1 - 2/2010 )

   = ( phần còn lại bạn tự tính nha )

k cho mình đó, bài này mình làm rồi nên đúng 100% lun, sorry nha mình ngại viết nhiều

a) \(K=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(\Leftrightarrow K=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)

\(\Leftrightarrow K=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(\Leftrightarrow K=2.\left(1-\frac{1}{2010}\right)\)

\(\Leftrightarrow K=2.\frac{2009}{2010}=\frac{2009}{1005}\)

b)  F=1/18 + 1/54 + 1/108 +...+ 1/990

=> \(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(\Leftrightarrow F=3.\left(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\right)\)

\(\Leftrightarrow F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)

\(\Leftrightarrow F=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)

\(\Leftrightarrow F=1-\frac{1}{33}=\frac{32}{33}\)

\(a,\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{16.18}+\frac{4}{18.20}\)

\(=\frac{4}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=2.\frac{9}{20}\)

\(=\frac{9}{10}\)

\(b,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

thank you

a, \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+..+\frac{4}{16.18}+\frac{4}{18.20}\)

\(=\frac{4}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{16}-\frac{1}{18}+\frac{1}{18}-\frac{1}{20}\right)\)

\(=2\cdot\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=2\cdot\frac{9}{20}=\frac{9}{10}\)

b, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)