câu A 4/2.5+4/5.8+.....+4/47.50
K
Khách
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DD
Đoàn Đức Hà
Giáo viên
25 tháng 8 2021
\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{47.50}\)
\(=\frac{5}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{47.50}\right)\)
\(=\frac{5}{3}\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{50-47}{47.50}\right)\)
\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{47}-\frac{1}{50}\right)\)
\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{4}{5}\)
TT
1
5 tháng 6 2023
Q=(-1)+(-3)+(-5)+...+(-99)
Dãy số trên là dãy số cách đều -4 đơn vị và có 51 số hạng.
\(\Rightarrow\) Q = [ -99 + ( -1) . 51 : 2 = -2550
Vậy Q= -2500
S= \(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.10}\) + ... + \(\dfrac{1}{47.50}\)
S= \(\dfrac{1}{3}\) . ( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\) + ... + \(\dfrac{3}{47.50}\) )
S= \(\dfrac{1}{3}\) . ( \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\) )
S = \(\dfrac{1}{3}\) . \(\dfrac{12}{25}\)
S= \(\dfrac{4}{25}\)
Vậy S = \(\dfrac{4}{25}\)
30 tháng 3 2018
a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=1-\frac{1}{25}\)
\(=\frac{24}{25}\)
30 tháng 3 2018
đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)
\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)
\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)
\(2A=\frac{1}{9}-\frac{1}{45}\)
\(2A=\frac{4}{45}\)
\(A=\frac{4}{45}\div2\)
\(A=\frac{2}{45}\)
DT
2
YV
9 tháng 3 2016
Ta có: 3S = 3/2.5 + 3/5.8 + ... + 3/47.50
3S = 1/2 - 1/5 + 1/5 - 1/8 + ... +1/47 - 1/50
3S = 1/2 - 1/50
3S = 12/25
=> S = 12/25 : 3 = 4/25
ML
9 tháng 3 2016
k, đây là dạng toán sai phân hữu hạn.
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số hạng tổng quát là 1/[n.(n+3)] = (1/3).[(n+3)-n]/[n.(n+3)] = (1/3). [1/n - 1/(n+3)]
=>
A = (1/3).[(1/2 - 1/5) + (1/5 - 1/8) + (1/8 - 1/11) +...+(1/44 - 1/47) + (1/47 - 1/50)]
= (1/3).[1/2 - 1/50]
= (1/3). (24/50) = (1/3).(12/25) = 4/25
vậy A = 4/25
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good luck!
VN
Võ Ngọc Phương
VIP
25 tháng 8 2023
Sửa đề:
\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)
\(A=4.\dfrac{33}{68}\)
\(A=\dfrac{33}{17}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 8 2023
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))
A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)
24 tháng 7 2021
Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)
\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)
3 tháng 8 2016
\(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{47.50}\)
\(\Rightarrow\frac{4}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{47.50}\right)\)
\(\Rightarrow\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+...+\frac{1}{47}-\frac{1}{50}\right)\)
\(\Rightarrow\frac{4}{3}\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(\Rightarrow\frac{4}{3}.\frac{9}{50}=\frac{6}{25}\)
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+...+ \(\dfrac{4}{47.50}\)
A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{47.50}\))
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\)+...+ \(\dfrac{1}{47}\) - \(\dfrac{1}{50}\))
A = \(\dfrac{4}{3}\).( \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\))
A = \(\dfrac{4}{3}\). \(\dfrac{24}{50}\)
A = \(\dfrac{16}{25}\)