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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
\(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{440}\)
= \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{20.22}\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{20}-\frac{1}{22}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{22}\right)=\frac{1}{2}.\frac{5}{11}=\frac{5}{22}\)
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.
TA
6 tháng 3 2019
Gọi A=\(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)
\(2A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{12}\)
\(2A=1-\frac{1}{12}=\frac{11}{12}\)
\(A=\frac{11}{12}:2=\frac{11}{24}\)
HV
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 11 2023
8 = 2 \(\times\) 4
24 = 4 \(\times\) 6
48 = 6 \(\times\) 8
80 = 8 \(\times\) 10
Xét dãy số: 2; 4; 6; 8;...; đây là dãy số cách đều với khoảng cách là:
4 - 2 = 2
Số thứ 20 của dãy số trên là: 2 x (20 - 1) + 2 = 40
Vậy Phân số thứ 20 của dãy số đã cho là: \(\dfrac{1}{40\times42}\)
Tổng của 20 phân số đầu tiên của dãy số đã cho là:
A = \(\dfrac{1}{8}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{48}\) + \(\dfrac{1}{80}\) +...+ \(\dfrac{1}{1680}\)
A = \(\dfrac{1}{2\times4}\) + \(\dfrac{1}{4\times6}\) + \(\dfrac{1}{6\times8}\) + \(\dfrac{1}{8\times10}\)+...+ \(\dfrac{1}{40\times42}\)
A = \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{2\times4}\) + \(\dfrac{2}{4\times6}\)+\(\dfrac{2}{6\times8}\)+\(\dfrac{2}{8\times10}\)+...+\(\dfrac{2}{40\times42}\))
A = \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{10}\)+...+ \(\dfrac{1}{40}\) - \(\dfrac{1}{42}\))
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{42}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{40}{42}\)
A = \(\dfrac{5}{21}\)
BC
2
19 tháng 7 2016
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}+\frac{1}{168}\)
\(=\frac{1}{4}.\left(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=\frac{1}{4}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{4}.\left(1+1-\frac{1}{7}\right)\)
\(=\frac{1}{4}.\left(2-\frac{1}{7}\right)\)
\(=\frac{1}{4}.\frac{13}{7}=\frac{13}{28}\)
\(D=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+........+\frac{1}{1520}\)
\(D=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+........+\frac{1}{38.40}\)
\(2D=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+......+\frac{2}{38.40}\)
\(2D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-.......-\frac{1}{40}\)
\(2D=1-\frac{1}{40}\)
\(2D=\frac{40}{40}-\frac{1}{40}\)
\(2D=\frac{39}{40}\)
\(D=\frac{39}{40}:2=\frac{39}{40}.\frac{1}{2}=\frac{39}{80}\)
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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