\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009+2011}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\frac{2010}{2011}\)

\(=\frac{1005}{2011}\)

tớ làm câu b thôi, câu a nhân 1/2 lên là đc 

\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)

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

\(A=\frac{1}{2}.\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2.\left(2n+1\right)}< \frac{1}{2}\)

p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)

 nhung ma ko cothoi gian giai

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)

làm tắt thế ai mà hỉu đc

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{3}.\frac{98}{99}\)

\(=\frac{98}{297}\)

Chuc bn học tốt

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

Đặt tổng là M

Ta có

\(M=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow M=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(\Rightarrow M=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)

=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)

=\(1-\frac{1}{2011}\)

=\(\frac{2010}{2011}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\)

\(=\frac{1\cdot2}{2\cdot1\cdot3}+\frac{1\cdot2}{2\cdot3\cdot5}+\frac{1\cdot2}{2\cdot5\cdot7}+...+\frac{1\cdot2}{2\cdot2009\cdot2011}\)

\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2009\cdot2011}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{2011}\right)\)= .......

Mình không chắc là đúng đâu nha

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\Leftrightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(\Leftrightarrow2A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}\)

\(\Leftrightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(\Leftrightarrow2A=1-\frac{1}{99}\)

\(\Leftrightarrow2A=\frac{99}{99}-\frac{1}{99}\)

\(\Leftrightarrow2A=\frac{98}{99}\)

\(\Leftrightarrow A=\frac{98}{99}\div2\)

\(\Leftrightarrow A=\frac{49}{99}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97+99}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

\(A=\left(1-\frac{1}{99}\right)+\left(-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}\right)\)

\(A=\frac{98}{99}+0\)

\(A=\frac{98}{99}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)

\(A=\frac{1}{2}.\frac{98}{99}\)

\(A=\frac{98}{198}=\frac{49}{99}\)

 M = 1 - 1/ 999 = 998/999

=> 2M = 2/1*3 + 2/3*5 + ... + 2/995*997  + 2/ 997*999  =  1-1/3 + 1/3 - 1/5 +... + 1/ 995 - 1/997 + 1/997 - 1 / 999 = 1- 1/999 = 998/999 

=> m = 998/999   /  2   =  499/999  (bn tính lại xem nha mk ko có máy tính nên sợ sai)

 Vậy M = ..... 

\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{995.997}+\frac{1}{997.999}\)

\(M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{995}-\frac{1}{997}+\frac{1}{997}-\frac{1}{999}\)

\(M=1-\frac{1}{999}\)

\(M=\frac{998}{999}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)

\(A=\frac{1}{2}.\frac{2012}{2013}\)

\(A=\frac{1006}{2013}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)

\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)

\(A=\frac{1}{2}.\frac{2012}{2013}\)

\(A=\frac{1006}{2013}\)

A = 1/1. x 1/3 + 1/3 x 1/5 + 1/5 x 1/7 x.....+1/2011 x 1/2013

A=1/1 x 1/2013(bạn triệt tiêu 1/3 ,1/5 ,1/7,1/2011 nha )

A =1/2013