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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Leftrightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(\Leftrightarrow2A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}\)
\(\Leftrightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(\Leftrightarrow2A=1-\frac{1}{99}\)
\(\Leftrightarrow2A=\frac{99}{99}-\frac{1}{99}\)
\(\Leftrightarrow2A=\frac{98}{99}\)
\(\Leftrightarrow A=\frac{98}{99}\div2\)
\(\Leftrightarrow A=\frac{49}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97+99}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)
\(A=\left(1-\frac{1}{99}\right)+\left(-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}\right)\)
\(A=\frac{98}{99}+0\)
\(A=\frac{98}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được:
\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)
\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)
=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)
\(4A>1=>A>\frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\frac{98}{99}\)
\(=\frac{98}{297}\)
Chuc bn học tốt
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)
=\(1-\frac{1}{2011}\)
=\(\frac{2010}{2011}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\)
\(=\frac{1\cdot2}{2\cdot1\cdot3}+\frac{1\cdot2}{2\cdot3\cdot5}+\frac{1\cdot2}{2\cdot5\cdot7}+...+\frac{1\cdot2}{2\cdot2009\cdot2011}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{2011}\right)\)= .......
Mình không chắc là đúng đâu nha
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> 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 = .....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\frac{2012}{2013}\)
\(A=\frac{1006}{2013}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\frac{2012}{2013}\)
\(A=\frac{1006}{2013}\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{35.37}\)
\(C=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{35.37}\right)\)
\(C=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{35}-\frac{1}{37}\right)\)
\(C=\frac{1}{2}.\left(1-\frac{1}{37}\right)\)
\(C=\frac{1}{2}.\frac{36}{37}\)
\(C=\frac{18}{37}\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{35.37}\)
\(C=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{35}-\frac{1}{37}\right)\)
\(C=\frac{1}{2}\cdot\left(1-\frac{1}{37}\right)\)
\(C=\frac{1}{2}\cdot\frac{36}{37}=\frac{18}{37}\)
Vay C = \(\frac{18}{37}\)