Các bạn giải giúp mình với :
5/2.5 + 5/5.8 + 5/8.11 +.....+ 5/98.101 =
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27 tháng 7 2018
\(\frac{5}{2.5}+\frac{5}{5.8}+......+\frac{5}{98.101}\)
\(=\frac{5}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+........+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)=\frac{5}{3}.\frac{99}{202}\)
\(=\frac{5.33}{202}=\frac{165}{202}\)
15 tháng 8 2016
\(S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\)
\(S=\frac{1}{2}-\frac{1}{101}\)
\(S=\frac{99}{202}\)
DD
Đoàn Đức Hà
Giáo viên
25 tháng 8 2021
\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{47.50}\)
\(=\frac{5}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{47.50}\right)\)
\(=\frac{5}{3}\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{50-47}{47.50}\right)\)
\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{47}-\frac{1}{50}\right)\)
\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{4}{5}\)
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
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)
\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\)
\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\)
\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)
24 tháng 5 2020
\(M= \dfrac{3^2}{2.5} +\dfrac{3^2}{5.8} +\dfrac{3^2}{8.11}+...+\dfrac{3^2}{98.101}\)
\(M= \) \( \dfrac{9}{2.5} +\dfrac{9}{5.8} +\dfrac{9}{8.11}+...+\dfrac{9}{98.101}\)
\(M=3(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+ \dfrac{3}{98.101})\)
\(M= 3(\dfrac{1}{2} -\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11})\)
\(M= 3(\dfrac{1}{2}-\dfrac{1}{11})\)
\(M=3(\dfrac{11}{22}- \dfrac{2}{22})\)
\(M=3.\dfrac{9}{22}\)
\(M=\dfrac{27}{22}\)
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
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)
NA
9
NB
27 tháng 4 2017
= 1/3.(1/2-1/5)+1/3.(1/5-1/8)+....+1/3.(1/92-1/95)+1/3.(1/95-1/98)
=1/3.(1/2-1/5+1/5-1/8+....+1/92-1/95+1/95-1/98)
=1/3.(1/2-1/98)
=1/3.24/49
=8/49
T
27 tháng 4 2017
Phân tích: 1/2.5 = 1/2 - 1/5
1/5.8 = 1/5 - 1/8
1/8.11 = 1/8 - 1/11
...
1/92.95 = 1/92 - 1/95
1/95.98 = 1/95 - 1/98
Ta có: 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
3 = 3/2.5 + 3/5.8 + 3/8.11 + ...+ 3/92.95 + 3/95.98
3 = 1 - 1/2 + 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
= 1 - 1/98
= 97/98 : 3 = 97/98 x 1/3 = (tự tính)
19 tháng 4 2020
a) e chỉ cần nhân chúng lại với nhau = cách tách từng cái ra
b)đặt 4/2.5+4/5.8+4/8.11+......+4/62.65 là S
\(.S=\frac{4}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{62.65}\right)\)
\(S=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{62}-\frac{1}{65}\right)\)
\(S=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{65}\right)\)
\(S=\frac{4}{3}\left(\frac{65}{130}-\frac{2}{130}\right)\)
\(S=\frac{4}{3}\left(\frac{63}{130}\right)\)
\(S=\frac{42}{65}\)
Bài 2 : \(\frac{15+a}{29+a}=\frac{3}{5}\)\(\Leftrightarrow\left(15+a\right)5=\left(29+a\right)3\Leftrightarrow75+5a=87+3a\Leftrightarrow5a-3a=87-75\Rightarrow2a=12\Rightarrow a=6\)
vậy a =6
\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{98.101}\)
\(=\frac{5}{2}-\frac{5}{5}+\frac{5}{5}-\frac{5}{8}+....+\frac{5}{98}-\frac{5}{101}\)
\(=\frac{5}{2}-\frac{5}{101}=\frac{495}{202}\)
\(\frac{5}{2\times5}+\frac{5}{5\times8}+\frac{5}{8\times11}+...+\frac{5}{98\times101}\)
\(=\frac{5}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{98\times101}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{5}{3}\times\frac{99}{202}=\frac{165}{202}\)