ST
2
K
Khách
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KK
2
ML
22 tháng 7 2017
\(K=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+...+\frac{10}{502.507}\)
\(\Leftrightarrow K=\frac{10}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+...+\frac{1}{502}-\frac{1}{507}\right)\)
\(\Leftrightarrow K=2\left(\frac{1}{7}-\frac{1}{507}\right)\)
\(\Leftrightarrow K=2\cdot\frac{500}{3549}\)
\(\Leftrightarrow K=\frac{1000}{3549}\)
6 tháng 2 2016
\(\Rightarrow C=\frac{10}{5}\left(\frac{1}{7.12}+\frac{1}{12.17}+\frac{1}{17.22}+...+\frac{1}{502.507}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{507}-\frac{1}{507}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{1}{7}-2.\frac{1}{507}=\frac{2}{7}-\frac{2}{507}\)
17 tháng 3 2017
D = \(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{2006.2009}\)
= \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{2006}-\dfrac{1}{2009}\)
= \(\dfrac{1}{5}-\dfrac{1}{9}=\dfrac{2004}{10045}\)
17 tháng 3 2017
C = \(\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\)
= \(\dfrac{10}{5}\left(\dfrac{5}{7.12}+\dfrac{5}{12.17}+\dfrac{5}{17.22}+...+\dfrac{5}{502.507}\right)\)
= \(\dfrac{10}{5}\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+....+\dfrac{1}{502}-\dfrac{1}{507}\right)\)
= \(\dfrac{10}{5}\left(\dfrac{1}{5}-\dfrac{1}{507}\right)\)
= \(\dfrac{10}{5}.\dfrac{502}{2535}\)
= \(\dfrac{1000}{3549}\)
NP
2
4 tháng 8 2020
\(A=\frac{1}{7\cdot12}+\frac{1}{12\cdot17}+\frac{1}{17\cdot22}+...+\frac{1}{52\cdot57}\)
\(A=\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+...+\frac{5}{52\cdot57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{52}-\frac{1}{57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{57}\right)=\frac{1}{5}\cdot\frac{50}{399}=\frac{10}{399}\)
\(B=\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+...+\frac{10}{253\cdot258}\)
\(B=\frac{10}{5}\left(\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+...+\frac{5}{253\cdot258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{258}\right)=2\cdot\frac{125}{1032}=\frac{125}{516}\)
4 tháng 8 2020
*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
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 = \(2015+\frac{10}{2,7}+\frac{10}{7,12}+\frac{10}{12,17}+\frac{10}{17,22}=2021.510611\)
TN
0
WH
18 tháng 3 2018
Sửa:
\(B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}\)
Trả lời
\(B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+\frac{5}{22\cdot27}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{27}\)
\(\Rightarrow B=\frac{25}{54}\)
Vậy B=\(\frac{25}{54}\)
11 tháng 8 2018
\(\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+\frac{5}{22\cdot27}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}\)
\(=\frac{1}{2}-\frac{1}{27}\)
\(=\frac{25}{54}\)
\(B=2\left(\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{502.507}\right)\)
\(B=2\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{502}-\frac{1}{507}\right)\)
\(B=2\left(\frac{1}{7}-\frac{1}{507}\right)\)
\(B=2\times\frac{500}{3549}\)
\(B=\frac{1000}{3549}\)
\(B=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+......+\frac{10}{502.507}\)
\(B=\frac{10}{5}.\left(\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+........+\frac{5}{502.507}\right)\)
\(B=\frac{10}{5}.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+.....+\frac{1}{502}-\frac{1}{507}\right)\)
\(B=\frac{10}{5}.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{500}{3549}=\frac{1000}{3549}\)