NP
2
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
ST
2
TN
13 tháng 5 2016
\(B=2\left(\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{502.507}\right)\)
\(B=2\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{502}-\frac{1}{507}\right)\)
\(B=2\left(\frac{1}{7}-\frac{1}{507}\right)\)
\(B=2\times\frac{500}{3549}\)
\(B=\frac{1000}{3549}\)
HP
13 tháng 5 2016
\(B=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+......+\frac{10}{502.507}\)
\(B=\frac{10}{5}.\left(\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+........+\frac{5}{502.507}\right)\)
\(B=\frac{10}{5}.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+.....+\frac{1}{502}-\frac{1}{507}\right)\)
\(B=\frac{10}{5}.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{500}{3549}=\frac{1000}{3549}\)
YK
17 tháng 1 2015
Nếu ai có giải dùm mình thì giải từng phần nhưng đừng chỉ ghi kết quả nhé~
CN
20 tháng 1 2016
a,\(\frac{2004}{10045}\)
b,\(\frac{25}{609}\)
c,\(\frac{1000}{3549}\)
d,\(\frac{25}{258}\)
KK
2
ML
22 tháng 7 2017
\(K=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+...+\frac{10}{502.507}\)
\(\Leftrightarrow K=\frac{10}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+...+\frac{1}{502}-\frac{1}{507}\right)\)
\(\Leftrightarrow K=2\left(\frac{1}{7}-\frac{1}{507}\right)\)
\(\Leftrightarrow K=2\cdot\frac{500}{3549}\)
\(\Leftrightarrow K=\frac{1000}{3549}\)
CB
2
MV
3 tháng 11 2017
\(D=\dfrac{4}{8\cdot13}+\dfrac{4}{13\cdot18}+\dfrac{4}{18\cdot23}+...+\dfrac{4}{253\cdot258}\\ =\dfrac{4}{5}\cdot\dfrac{5}{8\cdot13}+\dfrac{4}{5}\cdot\dfrac{5}{13\cdot18}+\dfrac{4}{5}\cdot\dfrac{5}{18\cdot23}+...+\dfrac{4}{5}\cdot\dfrac{5}{253\cdot258}\\ =\dfrac{4}{5}\left(\dfrac{5}{8\cdot13}+\dfrac{5}{13\cdot18}+\dfrac{5}{18\cdot23}+...+\dfrac{5}{253\cdot258}\right)\\ =\dfrac{4}{5}\cdot\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\\ =\dfrac{4}{5}\cdot\left(\dfrac{1}{8}-\dfrac{1}{258}\right)\\ =\dfrac{4}{5}\cdot\dfrac{125}{1032}\\ =\dfrac{25}{258}\)
LN
3 tháng 11 2017
ta có
Tính:
\(\dfrac{4}{8.13}+\dfrac{4}{13.18}+....+\dfrac{4}{253.258}\)
= 4 (\(\dfrac{1}{8.13}+\dfrac{1}{13.18}+.....+\dfrac{1}{253.258}\))
=\(\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
=\(\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)\)
=\(\dfrac{25}{258}\)
CN
3
DN
14 tháng 6 2015
\(N=2015+\frac{10}{2.7}+\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}\)
\(=2\left(1007,5+\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}\right)\)
\(=2\left(1007,5+\frac{7-2}{2.7}+\frac{12-7}{7.12}+\frac{17-12}{12.17}+\frac{22-17}{17.22}\right)\)
\(=2\left(1007,5+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}\right)\)
\(=2\left(1007,5+\frac{1}{2}-\frac{1}{22}\right)\)
\(=2015+1-\frac{1}{11}\)
\(=\frac{22175}{11}\)
CL
14 tháng 6 2015
N = \(2015+\frac{10}{2,7}+\frac{10}{7,12}+\frac{10}{12,17}+\frac{10}{17,22}=2021.510611\)
6 tháng 2 2016
\(\Rightarrow C=\frac{10}{5}\left(\frac{1}{7.12}+\frac{1}{12.17}+\frac{1}{17.22}+...+\frac{1}{502.507}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{507}-\frac{1}{507}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{1}{7}-2.\frac{1}{507}=\frac{2}{7}-\frac{2}{507}\)
9 tháng 5 2019
\(B=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(B=2\left[\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{28}\right]=\frac{25}{42}\)
NH
21 tháng 4 2019
\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)
\(A=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2.\frac{25}{84}=\frac{25}{42}\)
F
21 tháng 4 2019
\(A=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(A=10\left(\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+\frac{1}{13\cdot18}+\frac{1}{18\cdot23}+\frac{1}{23\cdot28}\right)\)
\(A=\frac{10}{5}\left(\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2\cdot\frac{25}{84}\)
\(A=\frac{25}{42}\)
\(A=\frac{1}{7\cdot12}+\frac{1}{12\cdot17}+\frac{1}{17\cdot22}+...+\frac{1}{52\cdot57}\)
\(A=\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+...+\frac{5}{52\cdot57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{52}-\frac{1}{57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{57}\right)=\frac{1}{5}\cdot\frac{50}{399}=\frac{10}{399}\)
\(B=\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+...+\frac{10}{253\cdot258}\)
\(B=\frac{10}{5}\left(\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+...+\frac{5}{253\cdot258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{258}\right)=2\cdot\frac{125}{1032}=\frac{125}{516}\)
*Cái đây giải thích hơi bị " khó hiểu " :
Chỗ mẫu (12 - 7) = (17 - 12) = ... = (57 - 52) = 5
Tử là 1 , mẫu là 5 nên tử/mẫu = 1/5
Hay \(\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+...+\frac{5}{52\cdot57}\right)\)
Còn bạn Trương Bùi Linh thì :
Mẫu = (13 - 8) = (18 - 13) = (23 - 18) = ... = 5
Tử là 10,mẫu là 5 => tử / mẫu = 10/5 = 2