tính \(B=\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}.+....+\frac{1}{301.304}\)
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
K
4
4 tháng 8 2016
A=\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\)
3A=3(\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\))
3A=\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
3A=\(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}+\frac{1}{76}\)
3A=\(\frac{1}{4}-\frac{1}{76}\)
3A=\(\frac{9}{38}\)
A=\(\frac{9}{38}\):3
A=\(\frac{3}{38}\)
4 tháng 8 2016
đặt A=1/4.7+1/7.10+...+1/73.76
3A=1/4-1/7+1/7-1/10+...+1/ 73 -1/ 76
27 tháng 4 2016
Ta thấy: 1/1-1/4 = 3/4 = 3.(1/1.4)
1/4-1/7 = 3/28 = 3.(1/4.7)
A = 3(1/1-1/4+1/4-1/7+...+1/97-1/100)
A = 3.(1-1/100)
A = 3.(99/100)
A = 297/100
27 tháng 4 2016
\(A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\frac{99}{100}\)
\(A=\frac{33}{100}\)
22 tháng 7 2015
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)
\(A=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)
\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\cdot\frac{97}{303}=\frac{97}{606}\)
\(B=\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+...+\frac{2}{100\cdot103}\)
\(B=\frac{2}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{100\cdot103}\right)\)
\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{103}\right)=\frac{2}{3}\cdot\frac{99}{412}=\frac{33}{206}\)
DP
29 tháng 6 2017
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)
ML
29 tháng 6 2017
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}.\frac{18}{19}\)
\(A=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)
\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}.\frac{5}{24}\)
\(B=\frac{5}{48}\)
28 tháng 6 2017
đây là toán lớp 5 cơ mà
a)A=\(\frac{1}{1x4}\)+\(\frac{1}{4x7}\)+...+\(\frac{1}{16x19}\)
A=\(\frac{1}{3}\)x3x(\(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+.......+\(\frac{1}{16.19}\)
A=\(\frac{1}{3}\)x(\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+............+\(\frac{3}{16.19}\))
A=\(\frac{1}{3}\)x(1-1/4+1/4-1/7+......+1/13-1/16+1/16-1/19)
A=\(\frac{1}{3}\)x(1-\(\frac{1}{19}\))
A=\(\frac{1}{3}\)x\(\frac{18}{19}\)
A=\(\frac{6}{19}\)
7 tháng 12 2016
http://olm.vn/hoi-dap/question/772291.html
sau 3 phút có kết quả tùy bạn
21 tháng 5 2016
a)b) Bạn nhân cả tử và mẫu với 2. Mình làm luôn, ko ghi lại đề bài
a)\(\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
=\(\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)
=\(\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)
b)\(\frac{2}{10.18}+\frac{2}{18.26}+\frac{2}{26.34}+...+\frac{2}{802.810}\)
=\(\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+\frac{1}{26}-\frac{1}{34}+...+\frac{1}{802}-\frac{1}{810}\right)\)
=\(\frac{1}{4}.\left(\frac{1}{10}-\frac{1}{810}\right)\)
=\(\frac{1}{4}.\frac{8}{81}=\frac{2}{81}\)
c) Mình biết làm, ddoiwtj tí nữa mình làm cho. Giờ đang mỏi tay
Thẳng Nobita kun có chép bài thì đừng t..i..c..k cho nó
B = \(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\)
B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{301}-\frac{1}{304}\right)\)
B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)\)
B = \(\frac{1}{3}.\frac{75}{304}\)
B = \(\frac{25}{304}\)
\(B=\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{301.304}\right):3\)
\(\Rightarrow B=\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right):3\)
\(\Rightarrow B=\left(\frac{1}{4}-\frac{1}{304}\right):3\)
\(\Rightarrow B=\frac{75}{304}:3=\frac{25}{304}\)