Question

tính tổng các phân số sau:

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2003.2004}\)                      b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}\)                  

Answer 1

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

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

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

                \(=1-\frac{1}{2005}\)

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

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

Answer 2

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

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

= \(1-\frac{1}{2005}\)

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

Answer 3

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 !!

Answer 4

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

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

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

Answer 5

Bổ sung phần b:

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