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24 tháng 5 2020
\(M= \dfrac{3^2}{2.5} +\dfrac{3^2}{5.8} +\dfrac{3^2}{8.11}+...+\dfrac{3^2}{98.101}\)
\(M= \) \( \dfrac{9}{2.5} +\dfrac{9}{5.8} +\dfrac{9}{8.11}+...+\dfrac{9}{98.101}\)
\(M=3(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+ \dfrac{3}{98.101})\)
\(M= 3(\dfrac{1}{2} -\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11})\)
\(M= 3(\dfrac{1}{2}-\dfrac{1}{11})\)
\(M=3(\dfrac{11}{22}- \dfrac{2}{22})\)
\(M=3.\dfrac{9}{22}\)
\(M=\dfrac{27}{22}\)
27 tháng 7 2018
\(\frac{5}{2.5}+\frac{5}{5.8}+......+\frac{5}{98.101}\)
\(=\frac{5}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+........+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)=\frac{5}{3}.\frac{99}{202}\)
\(=\frac{5.33}{202}=\frac{165}{202}\)
SK
1
QD
26 tháng 4 2017
Đề hình như bị sai ban ơi sửa lại
\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{95}\)
\(A=\dfrac{93}{190}\)
\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)
\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)
\(3B=2.\dfrac{93}{190}\)
\(3B=\dfrac{93}{95}\)
\(\Rightarrow B=\dfrac{31}{95}\)
HH
2
F
5 tháng 10 2020
\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{98.101}\)
\(=\frac{5}{2}-\frac{5}{5}+\frac{5}{5}-\frac{5}{8}+....+\frac{5}{98}-\frac{5}{101}\)
\(=\frac{5}{2}-\frac{5}{101}=\frac{495}{202}\)
5 tháng 10 2020
\(\frac{5}{2\times5}+\frac{5}{5\times8}+\frac{5}{8\times11}+...+\frac{5}{98\times101}\)
\(=\frac{5}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{98\times101}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{5}{3}\times\frac{99}{202}=\frac{165}{202}\)
PH
2
NH
26 tháng 2 2023
a) đặt
\(S=1+\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\\ 2S=2+\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\\ 2S=2+\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ 2S=2+1-\dfrac{1}{101}\\ 2S=\dfrac{302}{101}\\ S=\dfrac{151}{101}\)
b)
đặt
\(S=\dfrac{1}{2}+\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{98}-\dfrac{1}{101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{101}\\ 3S=\dfrac{201}{101}\\ S=\dfrac{67}{101}\)
TH
Thầy Hùng Olm
Manager
VIP
26 tháng 2 2023
\(2A-1=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(2A-1=1-\dfrac{1}{101}=\dfrac{100}{101}\)
\(2A=\dfrac{201}{101}\Rightarrow A=\dfrac{201}{202}\)
DC
3
7 tháng 8 2016
\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\frac{99}{202}\)
\(=\frac{66}{101}\)
LH
7 tháng 8 2016
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)
\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\)
\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\)
\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)
SK
2
T
8 tháng 6 2019
#)Giải :
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(\Rightarrow3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{99.101}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{101}\)
\(\Rightarrow3A=\frac{99}{202}\)
\(\Leftrightarrow A=\frac{33}{202}\)
8 tháng 6 2019
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)
11 tháng 4 2023
=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14
=1/2-1/14
=7/14-1/14=6/14=3/7
TT
0
\(S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\)
\(S=\frac{1}{2}-\frac{1}{101}\)
\(S=\frac{99}{202}\)
sai thì phải tử 3 sau tách ra thành tử 1