![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2B=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{23}-\frac{1}{25}\right)\)
\(2B=2\left(\frac{1}{3}-\frac{1}{25}\right)\)
\(2B=2\times\frac{22}{75}\)
\(B=\frac{44}{75}\)
![](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)
A=\(\dfrac{2}{1.3}-\dfrac{2}{3.5}-\dfrac{2}{5.7}-.....-\dfrac{2}{23.25}-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{23.25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{23}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\dfrac{22}{75}-\dfrac{1}{27}\)
A=\(\dfrac{227}{675}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M=(1/3-1/5)+(1/5+1/7)+...+(1/97+1/99)
M=1/3+(1/5-1/5)+...+(1/97-1/97)-1/99
M=1/3-1/99
M=32/99
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2/3.5+ 2 /5.7+ 2/7.9+...+ 2/97.99
=1/3 - 1/5 + 1/5 - 1 /7 +.... + 1/97 - 1/99
=1/3 - 1/99
=32/99
m=/3.5+2/5.7+2/7.9+.....+2/97.99
=m=1/3-1/5+1/5-1/7+.......+1/97-1/99
m=1/3-1/99
=32/99