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

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

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

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

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

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

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

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

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

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

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

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!!

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

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

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

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

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

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\)

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

                \(=1-\frac{1}{2005}\)

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

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

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

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

= \(1-\frac{1}{2004}\)

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

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

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

= \(1-\frac{1}{2005}\)

= \(\frac{2005}{2005}-\frac{1}{2005}=\frac{2004}{2005}\)

a, 1/ 1 . 2 + 1/2 . 3 + 1/3 . 4 + ... + 1/2003 . 2004

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2003 - 1/2004

= 1 - 1/2004

= 1 + ( -1 / 2004 )

= 2004 /2004 + ( -1 / 2004 )

= 2003 /2004

b, = 1/2 x ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + .... + 1/2003 - 1/2005

= 1/2 x ( 1 - 1/2005 )

= 1/2 x ( 2005 /2005 - 1/2005 )
= 1/2 x 2004/2005

= 1002 / 2005

Tíck nha !!

a) bạn xem lại đề nha

b)

\(B=\dfrac{1}{1.3}\)\(+\dfrac{1}{3.5}+...+\dfrac{1}{2003.2005}\)

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

\(B=\dfrac{1}{2}\left(1-\dfrac{1}{2005}\right)=\dfrac{1002}{2005}\)

A = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100

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

A= 1 - 1/100

A= 99/100

AXXXXXXXXXXXXXXXXXXXXXXX

ghi xong hết rồi

mạng nó rớt, ấn gửi trả lời mà không biết

tong teo

a)A = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100

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

Rút gọn ta được :

A= 1 - 1/100

A= 99/100

b) B = 1/1.3+1/3.5+1/5.7+....+1/ 99 .101

   B x 2 ta có : 1- 1/3 + 1/3 - 1/5+ 1/5-1/7+...+1/99-1/101

   B x2 rút gọn ta được: 1 - 1/ 101 

                         B x 2=  100 / 101

                        B =  100/ 101 : 2 = 50 / 101

a ) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{4}{20}+\frac{3}{20}\right)+\left(\frac{8}{21}+\frac{2}{21}-\frac{10}{21}\right)+\left(\frac{2}{5}-\frac{3}{5}\right)\)

\(=\frac{7}{20}+0+\frac{-1}{5}=\frac{7-4}{20}=\frac{3}{20}\)

b ) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{-21}{23}+\frac{-125}{143}\)

\(=0+\frac{-125}{143}=-\frac{125}{143}\)

bài 2

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

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

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

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

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

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

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

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