K = 4/2 - 4/4 + 4/4 - 4/6 + ....... + 4/2008 - 4/2010

K = 4/2 - 4/2010

K = 4016/2010 = 1/1003/1005

\(\frac{1004}{1005}\)

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

\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2002}-\frac{1}{2010}\right)\)

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

\(K=2.\frac{502}{1005}\)

\(K=\frac{1004}{1005}\)

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(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\))

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

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

A=\(\frac{1004}{1005}\)

Mình ko ghi lai đề nha

4/2.4/4+4/4.4/6+......+4/2008.4/2010=4/2.4/2010=4/1005

Mình ko bt đúng ko nữa nha 

A x 2/4= 2/2x4 + 2/4x6 + 2/6x8 +............+ 2/2008x2010

A x 2/4=4-2/2x4 + 6-4/4x6 + 8-6/6x8 +.......+ 2010-2008/2008x2010

A x 2/4=4/2x4 - 2/2x4 + 6/4x6 - 4/4x6 +8/6x8 -6/6x8 +............+ 2010/2008x2010 - 2008/2008x2010

A x 2/4=1/2-1/2010

A x 2/4=502/1005

A= 502/1005 / 2/4

A=1004/1005

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

\(C=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

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

\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\) \(;C=\frac{1}{2}.\frac{502}{1005}=\frac{251}{1005}\)

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

=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{1004.1005}\)

=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}\right)\)

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

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

=\(2.\frac{1004}{1005}\)

=\(\frac{2008}{1005}\)

P/s: Không biết đúng không nữa, làm đại ^.^

Ta thấy : \(\frac{4}{2.4}=\frac{1}{2}\left(\frac{4}{2}-\frac{4}{4}\right);\frac{4}{4.6}=\frac{1}{2}\left(\frac{4}{4}-\frac{4}{6}\right);...;\frac{4}{2008.2010}=\frac{1}{2}\left(\frac{4}{2008}-\frac{4}{2010}\right)\)

=> C =\(\frac{1}{2}.\left(\frac{4}{2}-\frac{4}{4}+\frac{4}{4}-\frac{4}{6}+\frac{4}{6}-\frac{4}{8}+...+\frac{4}{2008}-\frac{4}{2010}\right)\)

=> C = \(\frac{1}{2}\left(\frac{4}{2}-\frac{4}{2010}\right)=\frac{1}{2}\left(2-\frac{2}{1005}\right)=\frac{1}{2}\left(\frac{2010}{1005}-\frac{2}{1005}\right)\)

=> C = \(\frac{1}{2}\left(\frac{2010-2}{1005}\right)=\frac{1}{2}.\frac{2008}{1005}=\frac{1004}{1005}\)

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

=2.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\)

=2.(\(\frac{1}{2}-\frac{1}{2010}\))  =  2.(\(\frac{1005}{2010}-\frac{1}{2010}\))

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

=\(\frac{1004}{1005}\)

\(=2\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{1}{2010}\right)\)

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

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

\(=2\cdot\frac{1004}{2010}\)

\(=\frac{1004}{1005}\)

\(k\)\(mk\)\(nha\)\(bn\)

Gọi tổng đó là A Ta có

A:2=2/2x4+2/4x6+...+2/2008x2010

A:2=1/2-1/4+1/4-1/6+...+1/2008-1/2010

A:2=1/2-1/2010

A:2=1004/2010

A=1004/2010x2

A=2008/2010=1004/1005

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

\(\Rightarrow A=4\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2008.2010}\right)\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\right]\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)\right]\Rightarrow A=4\left(\frac{1}{2}.\frac{502}{1005}\right)\Rightarrow A=4.\frac{251}{1005}\Rightarrow A=\frac{1004}{1005}\)

\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)

\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)

\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)

\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)

= 2(2/2.4 + 2/4.6 +.....+ 2/2008.2016)

= 2(1/2 - 1/4 + 1/4 - 1/6 +....+ 1/2008 - 1/2016)

= 2(1/2 - 1/2016)

=2 . 1007/2016

=1007/1008

\(A=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)

\(A=2\left[\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right]\)

\(A=2\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right]\)

\(A=2\left[1-\frac{1}{2010}\right]=2\cdot\frac{2009}{2010}=\frac{2009}{1005}\)

A:2=2/2*4 + 2/4*6 + 2/6*8 + + ... + 2/2008*2010

A:2=1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2008 - 1/2010

A:2=1/2 - 1/2010

A:2=.,,( Bạn tự tính nhé)

các bạn khác chọn (k) đúng cho mình nhé

\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008+2010}\)

\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

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

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

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

\(=\frac{1004}{1005}\)

A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)

=\(7.\frac{3}{35}\)

=\(\frac{3}{5}\)

B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)

=\(\frac{1}{2}.\frac{2}{75}\)

=\(\frac{1}{75}\)

C : có ở bên dưới rồi, còn A và B thôi

1) A=7(\(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-......+\frac{1}{69}-\frac{1}{70}\) )

 A=7 ( \(\frac{1}{10}-\frac{1}{70}\))

A=7x 6/70

A=3/5

dễ mà bạn làm từ câu a nếu ra thì các câu khác cũng dễ thôi

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

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

\(A=\frac{2009}{2010}\)

Ta có:F=4/2.4+4/4.6+4/6.8+...+4/2008.2010

=4/2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)

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

=2.(1/2-1/2010)

=2.502/1005

=1004/1005

Mình chắc luôn đó, mình làm bài này rồi!