B=1.5+5.9+9.13+13.17+...+93.97+97.103
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CC
2
6 tháng 8 2016
\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)
6 tháng 8 2016
\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)
\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)
DT
0
DT
0
N
29 tháng 7 2018
a) \(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}=\frac{1}{5}\)
b) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}\)
\(=1-\frac{1}{17}=\frac{16}{17}\)
hok tốt ...
BT
29 tháng 7 2018
a)\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)
\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+\frac{2}{8\cdot10}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(A=\frac{2}{5}\cdot\frac{1}{2}=\frac{1}{5}\)
b)\(B=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)
AH
15 tháng 8 2016
- \(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\)
\(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)
\(4.B=1-\frac{1}{97}\)
\(4.B=\frac{96}{97}\)
\(B=\frac{96}{97}:4\)
\(B=\frac{24}{97}\)
NP
4
D
11 tháng 4 2017
\(4S=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\right)\)
=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}_{ }\)
=\(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{21}-\frac{1}{23}\)
=\(\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
=> \(S=\frac{4}{25}:4=\frac{4}{25}.\frac{1}{4}=\frac{1}{25}\)
11 tháng 4 2017
\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)
\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)
\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)
\(S\times4=\frac{1}{5}-\frac{1}{25}\)
\(S\times4=\frac{4}{25}\)
\(S=\frac{1}{25}\)
NP
2
25 tháng 2 2023
\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)
BF
25 tháng 2 2023
`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`
`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`
`= 1/5 - 1/25`
`= 4/25`
T
3
KT
15 tháng 8 2018
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
TM
15 tháng 8 2018
Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\) là A.
Ta có:
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)
\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)
\(4A=1-\frac{1}{45}\)
\(4A=\frac{44}{45}\)
\(A=\frac{44}{45}:4\)
\(A=\frac{11}{45}\)
Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)
S
10 tháng 5 2017
\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)
\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(=2100\left(1-\frac{1}{25}\right)\)
\(=2100\cdot\frac{24}{25}\)
\(=2016\)
10 tháng 5 2017
\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\frac{24}{25}\)
\(A=2016\)
TT
1
18 tháng 3 2019
\(A=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+...+\frac{2}{55\cdot59}\)
\(A=2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{55}-\frac{1}{59}\right)\)
\(A=2\left(\frac{1}{5}-\frac{1}{59}\right)\)
\(A=2\left(\frac{59}{295}-\frac{5}{295}\right)\)
\(A=2\cdot\frac{54}{295}\)
\(A=\frac{536}{295}\)
Sửa đề : 97.103 thành 97.101
B = 1.5 + 5.9 + 9.13 + 13.17 + ... + 93.97 + 97.101
12B = 1.5.12 + 5.9.12 + 9.13.12 + 13.17.12 + ... + 93.97.12 + 97.101.12
12B = 1.5.12 + 5.9.(13 - 1) + 9.13.(17-5) + ... + 97.101.(105-93)
12B = 1.5.12 + 5.9.13 - 1.5.9 + 9.13.17 - 5.9.13 + ... + 97.101.105 - 93.97.101
12B = 1.5.12 - 1.5.9 + 97.101.105
12B = 5 . (12 - 9) + 1028685
12B = 5 . 3 + 1028685
12B = 12 + 1028685
12B = 1028700
B = 85725