a) 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/2003.2004 = 1/1 - 1/2 +1/2 - 1/3 +...+ 1/2003 -1/2004 = 1 - 1/2004

b) Đặt B = 1/1.3 + 1/3.5 + 1/5.7 +...+ 1/2003.2005 => 2B = 2(1/1.3 + 1/3.5 + 1/5.7 +...+ 1/2003.2005) => 2B = 2/3.5 + 2/5.7 + 2/7.9 +...+ 2/2003.2005 => 2B = 1/3 - 1/5 + 1/5 - 1/7 +1/7 - 1/9 +...+ 1/2003 - 1/2005 => 2B = 1/3 - 1/2005 = 2012/6015 => B = 2012/6015 : 2 = 1001/6015

( Cái này là để bạn hiểu thêm cách mình làm ở trên : C/m : a/k.(k+a) = a/k - a/k+a

Ta có : a/k.(k+a) = (k+a) - k/k.(k+a) = k+a/k.(k+a) - k/k.(k+a) = a/k - a/k+a)

Bấm đúng cho mình nhe

sai rồi

mày bảo người ta làm sai thế mày làm đi . ooooooooooookkkkkkkkkkkk

chứ

a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2003.2004}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}=1-\dfrac{1}{2004}=\dfrac{2003}{2004}\)b)Đặt  \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)

\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)\(\Rightarrow A=\dfrac{1002}{2005}\)

a: Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\)

\(=\dfrac{2003}{2004}\)

b: Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2003\cdot2005}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2003\cdot2005}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2004}{2005}=\dfrac{1002}{2005}\)

Gọi \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2003.2004}\)

         \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)(tối giản các phân số giống nhau)

         \(A=\frac{1}{1}-\frac{1}{2004}\)

          \(A=\frac{2003}{2004}\)

gọi biểu thức trên là A, a có:

A=1/1.2+1/2.3+...+1/2003.2004

2A=2/1.2+2/2.3+...+2/2003.2004

2A=1/1-1/2+1/2-1/3+...+1/2003-1/2004

2A=1/1-1/2+1/2-1/3+...+1/2003-1/2004

2A=1/1-1/2004

2A=2003/2004

=>A=2003/2004:2

=>A=2003/4008

Bạn Nhi làm đúng rồi đấy! 

\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2003\times2004}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}=1-\frac{1}{2004}=\frac{2003}{2004}\)

1/1.2+1/2.3+1/4.4+...1/2003.2004

=1-1/2004

=2003/2004

a)1/1x2+1/2x3+....+1/2003x2004

=1-1/2+1/2-1/3+...+1/2003+1/2004

=1-1/2004

=2004/2004-1/2004

=2003/2004

b)1/1x3+1/3x5+...+1/2003x2005

=1-1/3+1/3-1/5+....+1/2003+1/2005

=1-1/2005

=2005/2005-1/2005

=2004/2005

2004/2005

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\)\(\frac{1}{2003.2004}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

=\(\frac{1}{1}-\frac{1}{2004}=\frac{2003}{2004}\)

b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\)\(\frac{1}{2003.2005}\)

=\(\frac{2}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\right)\)

=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2003.2005}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2005}\right)\)

=\(\frac{1}{2}.\frac{2004}{2005}\)

=\(\frac{1002}{2005}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2003}-\dfrac{1}{2004}=\dfrac{2003}{2004}\)

Ta có:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}=\frac{2003}{2004}\)

b,

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\right).\frac{1}{2}\)

\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\right).\frac{1}{2}\)

\(=\left(1-\frac{1}{2005}\right).\frac{1}{2}=\frac{2004}{2005}.\frac{1}{2}=\frac{1002}{2005}\)

Nhớ nha bạn

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}=\frac{2003}{2004}\)

b) Đặt A=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2003\cdot2005}\)

\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(2A=1-\frac{1}{2005}\)

\(2A=\frac{2004}{2005}\)

\(A=\frac{2004}{2005}:2=\frac{2004}{2005}\cdot\frac{1}{2}=\frac{1002}{2005}\)

a)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=\frac{1}{1}-\frac{1}{2004}\)

\(\Rightarrow=\frac{2003}{2004}\)

b)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003+2005}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(=\frac{1}{1}-\frac{1}{2005}\)

\(\Rightarrow=\frac{2004}{2005}\)

\(a,\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2003.2004}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2004}{2004}-\frac{1}{2004}=\frac{2003}{2004}\)

b) Đặt \(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)

\(\Rightarrow2B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\)

\(\Rightarrow2B=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(\Rightarrow2B=1-\frac{1}{2005}\)

\(\Rightarrow2B=\frac{2005}{2005}-\frac{1}{2005}\)

\(\Rightarrow2B=\frac{2004}{2005}\)

\(\Rightarrow B=\frac{2004}{2005}:2=\frac{2004}{2005}.\frac{1}{2}\)

\(\Rightarrow B=\frac{1002}{2005}\)

Vậy...

hok tốt!!

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2003\cdot2004}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2004}{2004}+\frac{-1}{2004}=\frac{2003}{2004}\)

                                    #Hoq chắc _ Baccanngon