3/3.5+3/5.7+...+3/97.99
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +....... + 1/97 - 1/99
= 1- 1/99
= 98/99
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
M=\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{95.97}+\dfrac{3}{97.99}\)
M=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\right)\)
M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
M=\(\dfrac{3}{2}.\dfrac{32}{99}\)
M=\(\dfrac{16}{33}\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{3}{3.5} + \dfrac{3}{5.7} + ... + \dfrac{3}{97.99}\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\dfrac{32}{99}\)
\(\Rightarrow A=\dfrac{16}{33}\)
Vậy \(A=\dfrac{16}{33}\)
A= \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{97.99}\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\dfrac{32}{99}\)
= \(\dfrac{3.32}{2.99}\)= \(\dfrac{3.2.3.6}{2.11.3.3}\)= \(\dfrac{6}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
2/3.C = 2/3.5 + 2/5.7 + .... + 2/97.99
= 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/97 - 1/99
= 1/3 - 1/99 = 32/99
=> C = 32/99 : 2/3 = 16/33
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=3.(3/1.3+3/3.5+3/5.7+...+3/95.97+3/97.99)
=3(1-1/3+1/3-1/5+1/5-1/7+...+1/95-1/97+1/97-1/99)
=3[(1-1/99)+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)]
=3(1-1/99)=3(99/99-1/99)=3.98/99=1.98/33=98/33
![](https://rs.olm.vn/images/avt/0.png?1311)
M = 1/3-1/5+1/5-1/7+...+1/97-1/99
= 1/3-1/99
=33/99-1/99
=32/99
M=1/2.(3/3-3/5+3/5-3/7+3/7-3/9+......+3/97-3/99)
M=1/2.(1-3/99)
M=1/2.32/33=16/33
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)
\(=\frac{3}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{32}{99}=\frac{16}{33}\)
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