Tính :
D= \(\frac{5}{1.4}\)+ \(\frac{5}{4.7}\)+ .....+\(\frac{5}{100.103}\)
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
TH
4
6 tháng 4 2017
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)
\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
6 tháng 4 2017
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
US
5
1 tháng 5 2016
\(\frac{5}{1.4}+\frac{5}{4.7}+................+\frac{5}{100.103}\)
\(\frac{1}{3}.\left(5-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+..............+\frac{5}{100}-\frac{5}{103}\right)\)
\(\frac{1}{3}.\left(5-\frac{5}{103}\right)\)
\(\frac{1}{3}.\left(\frac{510}{103}\right)=\frac{170}{103}\)
1 tháng 5 2016
\(=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{....}{....}\)
LT
0
NC
2
1 tháng 5 2016
=5/3.(1/1-1/4+1/4-1/7+...+1/100-1/103)
=5/3.(1/1-1/103)
=5/3.102/103
=170/103
29 tháng 5 2016
=5/3.(1/1-1/4+1/4-1/7+...+1/100-1/103)
=5/3.(1/1-1/103)
=5/3.102/103
=170/103
NT
1
22 tháng 4 2015
S = \(5-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+.......+\frac{5}{100}-\frac{5}{103}\)
S = \(5-\frac{5}{103}\)
S = \(\frac{510}{103}\)
21 tháng 10 2016
a cong tru loan nen ko hieu
b
A=5/1.4+5/4.7+..5/100.103
3/5.A=3/1.4+3/4.7+..+3/100.103
=1/1-1/4+1/4-1/7+...+1/100-1/103
=1-1/103=102/103
A=(5.102)/(3.103)=5.34/103
HT
2
Q
22 tháng 4 2017
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(3B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)
\(3B=5\left(1-\frac{1}{103}\right)\)
\(3B=5.\frac{102}{103}\)
\(3B=\frac{510}{103}\)
\(\Rightarrow B=\frac{170}{103}\)
22 tháng 4 2017
Ta có:
B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}+.....+\frac{5}{100.103}\)
B=\(\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{100.103}\right)\)
B=\(\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\right)\)
B=\(\frac{5}{3}\left(1-\frac{1}{103}\right)\)
B=\(\frac{5}{3}.\frac{102}{103}\)
B=\(\frac{170}{103}\)
Vậy B=\(\frac{170}{103}\)
nhớ k
AM
4 tháng 7 2015
a)\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}\cdot\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{170}{103}\)b)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\cdot\frac{16}{51}=\frac{8}{51}\)
5 tháng 5 2017
Câu a) bạn Ác Mộng làm rồi nên mình làm b) nha
b)Gọi A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(2A=\frac{1}{3}-\frac{1}{51}\)
\(2A=\frac{16}{51}\)
\(A=\frac{16}{51}:2\)
\(A=\frac{8}{51}\)
WH
6 tháng 5 2018
Trả lời
\(B=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+...+\frac{5}{100\cdot103}\)
\(\frac{3}{5}B=\frac{3}{5}\left(\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+...+\frac{5}{100\cdot103}\right)\)
\(\frac{3}{5}B=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{...3}{100\cdot103}\)
\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
\(\frac{3}{5}B=1-\frac{1}{103}\)
\(\frac{3}{5}B=\frac{102}{103}\)
\(B=\frac{102}{103}:\frac{3}{5}\)
\(B=\frac{170}{103}\)
6 tháng 5 2018
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)
\(3B=15\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(3B=15\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(3B=15\left(\frac{1}{1}-\frac{1}{100}\right)=15\left(\frac{100}{100}-\frac{1}{100}\right)=15.\frac{99}{100}\)
\(B=\frac{1}{3}.15-\frac{1}{3}.\frac{99}{100}=5-\frac{33}{100}=\frac{500}{100}-\frac{33}{100}=\frac{467}{100}\)
=\(\frac{5}{3}\cdot\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{100\cdot103}\right)\)
=\(\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
=\(\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)\)
=\(\frac{5}{3}\cdot\frac{102}{103}\)=\(\frac{170}{103}\)
Vậy D=\(\frac{170}{103}\)