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DT
0
DT
0
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
\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\frac{24}{25}\)
\(A=2016\)
21 tháng 4 2020
Chứng minh \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)
\(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)
\(A=\frac{1}{1}-\frac{1}{21}\)
\(A=\frac{20}{21}\)
\(\frac{20}{21}< 1\)
=> \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)( đpcm )
* Mình sợ sai xD *
TT
1
18 tháng 3 2019
\(A=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+...+\frac{2}{55\cdot59}\)
\(A=2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{55}-\frac{1}{59}\right)\)
\(A=2\left(\frac{1}{5}-\frac{1}{59}\right)\)
\(A=2\left(\frac{59}{295}-\frac{5}{295}\right)\)
\(A=2\cdot\frac{54}{295}\)
\(A=\frac{536}{295}\)
CC
1
VA
16 tháng 7 2016
Ta có :
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\)\(=1\)
\(x+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=1\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=1\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=1\)
\(x+\frac{8}{45}=1\)
\(x=1-\frac{8}{45}=\frac{37}{45}\)
Ủng hộ mk nha !!! ^_^
NT
2
15 tháng 4 2019
\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{37.41}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{37}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\frac{36}{205}=\frac{9}{205}\)
NH
15 tháng 4 2019
\(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{37.41}\)
\(=\frac{1}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{37.41}\right)\)
\(=\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{37}-\frac{1}{41}\right)\)
\(=\frac{1}{4}\left(\frac{1}{5}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\frac{36}{205}=\frac{9}{205}\)
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
\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)
\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)
\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)
\(S\times4=\frac{1}{5}-\frac{1}{25}\)
\(S\times4=\frac{4}{25}\)
\(S=\frac{1}{25}\)
NN
0
M
1