s = 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5

S=1 + (-1/2 +1/2)+...+(-1/4 + 1/4 ) +-1/5

S = 1 + 0 +0 +...+ 0 +-1/5

S= 1 + -1/5

S = 4/5

S=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5

S=1-1/5

S=4/5.

P=1/1.3+1/3.5+1/5.7+1/7.9

2P=2/1.3+2/3.5+2/5.7+2/7.9

2P=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

2P=1-1/9=8/9

P=8/9:2

P=4/9.

Chac chan dung nha ban.k cho minh nhe

S=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5 =1/1-1/5 nhé  P=1/2+(2/1.3+2/3.5+2/5.7+2/7.9)=1/2+(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9).=1/2+(1/1-1/9)

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) 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

S=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

\(S=\dfrac{1}{1}-\dfrac{1}{5}\\ S=\dfrac{4}{5}\)

\(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}\\ 2.P=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)

\(2.P=\dfrac{1}{1}-\dfrac{1}{9}\\ 2.P=\dfrac{8}{9}\\ P=\dfrac{8}{9}:2\\ P=\dfrac{8}{18}=\dfrac{4}{9}\)

còn cần không bạn, mk làm cho

A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)

=1\(-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)

=\(\dfrac{47}{60}\)

B=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)=

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}+\dfrac{1}{101}\)

=\(1-\dfrac{1}{101}\)

=\(\dfrac{100}{101}\)

A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)

=\(1-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)

= \(\dfrac{47}{60}\)

B= \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

= \(2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

= 2\(\left(1-\dfrac{1}{101}\right)\)

= \(\dfrac{200}{101}\)

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

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

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

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

\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)

A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)