Tính :
C=1/5.7+ 1/7.9+........+1/97.99.
Giải chi tiêt giúp mik ko làm tắt nha.
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
H
5 tháng 4 2016
Mk bik câu B nè!
2B = 2/3.5 + 2/5.7 + 2/7.9 +.......+2/97.99
2B = 1/3 - 1/5 + 1/5 - 1/7 +.......+ 1/97 - 1/99
2B = 1/3 - 1/99
2B = 32/99
=> B = 16/99
11 tháng 5 2019
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
TN
0
PH
1
CT
7
11 tháng 5 2018
\(M=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{1}{2}.\frac{32}{99}\)
\(\Rightarrow M=\frac{16}{99}\)
KR
11 tháng 5 2018
\(M=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5.}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{16}{99}\)
NP
1
VQ
30 tháng 3 2023
A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)
2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)
2A= 1-\(\dfrac{1}{99}\)
2A= \(\dfrac{98}{99}\)
A= \(\dfrac{98}{99}\) : 2
A=\(\dfrac{49}{99}\)
AN
2
15 tháng 8 2015
\(S=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}\)
\(S=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\right)\)
\(S=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}\cdot\frac{32}{99}\)
\(S=\frac{16}{99}\)
NP
1
S
30 tháng 3 2023
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(=\dfrac{49}{99}\)
DT
Đỗ Thanh Hải
CTVVIP
11 tháng 4 2021
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
\(C=\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{97\cdot99}\\ C=\dfrac{1}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{97\cdot99}\right)\\ C=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ C=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{99}\right)\\ C=\dfrac{1}{2}\cdot\dfrac{94}{495}\\ C=\dfrac{47}{495}\)
\(C=\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(C=\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(C=\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{99}\right)\)
\(C=\dfrac{1}{2}.\dfrac{94}{495}\)
\(C=\dfrac{47}{495}\)
\(#NqHahh\)