D=1/8+1/24+1/48+1/80+....=1/1520
K
Khách
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Những câu hỏi liên quan
12 tháng 2 2018
\(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{440}\)
\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+....+\frac{1}{20\cdot22}\)
\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+.....+\frac{2}{20\cdot22}\)
\(2A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{20}-\frac{1}{22}\)
\(2A=1-\frac{1}{22}\)
\(A=\frac{21}{22}:2\)
\(A=\frac{21}{44}\)
12 tháng 2 2018
\(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{440}\)
= \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{20.22}\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{20}-\frac{1}{22}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{22}\right)=\frac{1}{2}.\frac{5}{11}=\frac{5}{22}\)
LT
Lương Thị Vân Anh
CTVHS
2 tháng 3 2023
\(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(=1+\dfrac{1}{2\times4}+\dfrac{1}{4\times6}+\dfrac{1}{6\times8}+\dfrac{1}{8\times10}+\dfrac{1}{10\times12}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{12}=\dfrac{17}{12}\)
28 tháng 2 2018
A = 1 + 1/2.4 + 1/4.6 + 1/6.8 + 1/8.10 + 1/10.12
2A = 2 + 2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 + 2/10.12
= 2 + 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + 1/8 - 1/10 + 1/10 - 1/12
= 2 + 1/2 - 1/12 = 29/12
=> A = 29/12 : 2 = 29/24
Tk mk nha
KN
4
D
27 tháng 5 2022
`1/8+1/24+1/48+1/80+1/120`
`=1/[2xx4]+1/[4xx6]+1/[6xx8]+1/[8xx10]+1/[10xx12]`
`=1/2xx(2/[2xx4]+2/[4xx6]+2/[6xx8]+2/[8xx10]+2/[10xx12])`
`=1/2xx(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)`
`=1/2xx(1/2-1/12)`
`=1/2xx(6/12-1/12)`
`=1/2xx5/12=5/24`
27 tháng 5 2022
\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
=\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{10.12}\)
=\(\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{10.12}\right)\)
=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
=\(\dfrac{1}{2}.\dfrac{5}{12}\)
=\(\dfrac{5}{24}\)
Dấu chấm(.)là nhân.
NT
3
TT
5 tháng 5 2022
A=1+18+124+148+180+1120A=1+18+124+148+180+1120
=1+12.4+14.6+16.8+18.10+110.12=1+12.4+14.6+16.8+18.10+110.12
=1+12(12−14+14−16+16−18+18−110+110−112)=1+12(12−14+14−16+16−18+18−110+110−112)
=1+12(12−112)=1+12(12−112)
=1+524=1+524
=2924
Tham khảo thôi nka
LM
5 tháng 5 2022
2A= 2/8+2/24+2/48+2/80= 2/(2*4)+2/(4*6)+2/(6*8)+2/(8*10)= 1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10= 1/2-1/10= 2/5 =>A= 1/5
DD
Đoàn Đức Hà
Giáo viên
9 tháng 7 2021
\(A=\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\frac{1}{48}+\frac{1}{63}+\frac{1}{80}\)
\(=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)
\(=\frac{1}{2}\times\left[\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\right)+\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}\right)+\left(\frac{4-2}{2\times4}+\frac{6-4}{4\times6}+\frac{8-6}{6\times8}+\frac{10-8}{8\times10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)
\(=\frac{29}{45}\)
TD
1
5 tháng 4 2021
Ta có: \(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(\Leftrightarrow2A=2+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{24}{12}+\dfrac{6}{12}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{29}{12}\)
hay \(A=\dfrac{29}{24}\)
NT
4
NH
12 tháng 5 2019
\(\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\frac{1}{48}+\frac{1}{63}+\frac{1}{80}\)
\(=\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)+\left(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}\right)\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)+\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)
\(D=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+........+\frac{1}{1520}\)
\(D=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+........+\frac{1}{38.40}\)
\(2D=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+......+\frac{2}{38.40}\)
\(2D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-.......-\frac{1}{40}\)
\(2D=1-\frac{1}{40}\)
\(2D=\frac{40}{40}-\frac{1}{40}\)
\(2D=\frac{39}{40}\)
\(D=\frac{39}{40}:2=\frac{39}{40}.\frac{1}{2}=\frac{39}{80}\)
Vậy ....
\(D=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{1520}\)
\(D=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{38.40}\)
\(D=\frac{1}{2}\times\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{38.40}\right)\)
\(D=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{38}-\frac{1}{40}\right)\)
\(D=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{40}\right)\)
\(D=\frac{1}{2}\times\frac{19}{40}\)
\(D=\frac{19}{80}\)
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