Ta có: 

A =  \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\) + ... + \(\frac{2}{2001.2003}\) + \(\frac{2}{2003.2005}\)

    =  \(\frac{1}{1}\) -  \(\frac{1}{3}\)+ \(\frac{1}{3}\)-  \(\frac{1}{5}\)+ \(\frac{1}{5}\)-  \(\frac{1}{7}\)+  \(\frac{1}{7}\)-  \(\frac{1}{9}\) + ... + \(\frac{1}{2001}\)- \(\frac{1}{2003}\)+ \(\frac{1}{2003}\)-  \(\frac{1}{2005}\)

    = 1 - \(\frac{1}{2005}\)

     = \(\frac{2004}{2005}\)

Chúc bạn học tốt nha ^^!!

giúp mình với

Dễ mà bn , mình học dạng này òi

Trình bày cho mình nhé

mk có 3 cáh mn xem cáh nào hen

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

\(=\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(=\frac{100}{101}.\frac{3}{2}=\frac{105}{101}\)

c2 nhé

\(A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.......+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=3\left(1-\frac{1}{101}\right)=3.\frac{100}{101}=\frac{300}{101}\)

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)

\(3A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(3A=\frac{1}{1}-\frac{3}{101}\)\(\Rightarrow A=\left(1-\frac{1}{101}\right):3=\frac{100}{303}\)

                           \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

                 \(=\frac{2}{3}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

                  \(=\frac{2}{3}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

                   \(=\frac{2}{3}.\left(1-\frac{1}{51}\right)\)

                  \(=\frac{2}{3}.\frac{50}{51}=\frac{20}{51}\)

              Ủng hộ mk nha !!! ^_^

25/17 mới đúng

\(\frac{3}{1.3}\)+ \(\frac{3}{3.5}\)+ \(\frac{3}{5.7}\)+...+ \(\frac{3}{49.51}\)

= \(\frac{3}{2}\)( \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+...+ \(\frac{2}{49.51}\))

= \(\frac{3}{2}\)( \(\frac{1}{1}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+...+ \(\frac{1}{49}\)- \(\frac{1}{51}\))

= \(\frac{3}{2}\)( 1- \(\frac{1}{51}\))

= \(\frac{3}{2}\). \(\frac{50}{51}\)

= \(\frac{25}{17}\).

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +....... + 1/97 - 1/99

= 1- 1/99

= 98/99

Mãi yêu chàng trai Song Tử:bạn làm vầy là ngu roày

3/1.3+3/3.5+3/5.7+....+3/97.99=98/99

3/1.3 + 3/3.5 + 3/5.7 + ....... + 3/49.51

= 3 x ( 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/49.51 )

= 3 x ( 1 - 1/51 )

= 3 x      50/51

=       150/151

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

 
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)