K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
VT
15 tháng 4 2017
=1/2(2/3.5 + 2/5.7 +.....+2/49.51
=1/2(1/3 - 1/5+1/5-1/7+....+1/49-1/51)
=1/2(1/3-1/51)
=1/2.16/51
=8/51
HỌC TỐT NHÉ BẠN!
6 tháng 5 2018
\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(A=\frac{1}{2}.\frac{16}{51}\)
\(A=\frac{8}{51}\)
6 tháng 5 2018
\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\)
\(2A=\frac{1}{3}-\frac{1}{50}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{50}\right)\)
\(A=\frac{1}{2}.\frac{1}{3}-\frac{1}{2}.\frac{1}{50}\)
\(A=\frac{1}{6}-\frac{1}{100}=\frac{50}{300}-\frac{3}{300}=\frac{47}{300}\)
TH
1
10 tháng 4 2017
C=1/15+1/35+1/63+..+1/2499
=1/3.5+1/5.7+1/7.9+...+1/49.51
=1/2(2/3.5+2/5.7+2/7.9+...+2/49.51)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/49-1/51)
= 1/2.(1/3-1/51)
=1/2.16/51
=8/51
OB
0
4 tháng 3 2017
\(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\\ =\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}+\dfrac{1}{51}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)=\dfrac{1}{2}.\dfrac{16}{51}\\ =\dfrac{8}{51}\)
11 tháng 5 2019
\(E=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
TH
3
6 tháng 4 2017
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{49.51}\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
6 tháng 4 2017
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{8}{51}\)
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}.\dfrac{16}{51}=\dfrac{8}{51}\)
Vậy \(C=\dfrac{8}{51}\)
\(\Rightarrow C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(C=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(C=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(C=\dfrac{1}{2}.\dfrac{15}{51}\)
\(C=\dfrac{8}{51}\)