1/5.9+1/9.13+1/13.17+...+1/37.41=???
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Khách
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17 tháng 7 2015
A= 1/5.9+1/9.13+1/13.17+1/17.21+1/21.25
4A= 4/5.9+4/9.13+4/13.17+4/17.21+4/21.25
4A= (1/5-1/9)+(1/9-1/13)+(1/13-1/17)+(1/17-1/21)+(1/21-1/25)
4A= 1/5- 1/25
4A= 4/25
A= 4/25 :4
A= 1/25
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
\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)
\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)
\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)
\(S\times4=\frac{1}{5}-\frac{1}{25}\)
\(S\times4=\frac{4}{25}\)
\(S=\frac{1}{25}\)
ND
Nguyễn Đăng Nhân
CTVHS
20 tháng 2 2023
Tính chất của phân số bạn cần biết như sau:
\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)
Gọi biểu thức trên là A ,ta có:
\(A=\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+\dfrac{1}{13\cdot17}+\dfrac{1}{17\cdot21}+\dfrac{1}{21\cdot25}\)
\(4A=\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+\dfrac{4}{17\cdot21}+\dfrac{4}{21\cdot25}\)
\(4A=\dfrac{9-5}{5\cdot9}+\dfrac{13-9}{9-13}+\dfrac{17-13}{13\cdot17}+\dfrac{21-17}{17\cdot21}+\dfrac{25-21}{21\cdot25}\)
Áp dụng tính chất phân số đã nêu ở trên, ta được:
\(4A=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{25}\)
\(4A=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{5}{25}-\dfrac{1}{25}=\dfrac{4}{25}\)
\(A=4A:4=\dfrac{4}{25}:4=\dfrac{16}{25}\)
Vậy \(A=\dfrac{16}{25}\)
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 \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)
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 !!! ^_^
TH
3
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)\(B=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)
LD
1
24 tháng 8 2017
\(S1=\dfrac{5}{10.11}+\dfrac{5}{11.12}+.............+\dfrac{5}{14.15}\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+...............+\dfrac{1}{14.15}\right)\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+.............+\dfrac{1}{14}-\dfrac{1}{15}\right)\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{15}\right)\)
\(\Leftrightarrow S1=5.\dfrac{1}{30}=\dfrac{1}{6}\)
\(S2=\dfrac{1}{5.9}+\dfrac{1}{9.13}+\dfrac{1}{13.17}+........+\dfrac{1}{21.25}\)
\(\Leftrightarrow4S_2=\dfrac{4}{5.9}+\dfrac{4}{9.13}+..............+\dfrac{4}{21.25}\)
\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+............+\dfrac{1}{21}-\dfrac{1}{25}\)
\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{25}\)
\(\Leftrightarrow4S_2=\dfrac{4}{25}\)
\(\Leftrightarrow S_2=\dfrac{16}{25}\)
KB
8 tháng 4 2022
Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{37.41}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{37}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\frac{36}{205}=\frac{9}{205}\)
\(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{37.41}\)
\(=\frac{1}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{37.41}\right)\)
\(=\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{37}-\frac{1}{41}\right)\)
\(=\frac{1}{4}\left(\frac{1}{5}-\frac{1}{41}\right)\)
\(=\frac{1}{4}.\frac{36}{205}=\frac{9}{205}\)