ND
2
K
Khách
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2
7 tháng 2 2023
`1/15+1/35+1/63+1/99+1/143`
`=1/[3.5]+1/[5.7]+1/[7.9]+1/[9.11]+1/[11.13]`
`=1/2(2/[3.5]+2/[5.7]+2/[7.9]+2/[9.11]+2/[11.13])`
`=1/2.(1/3-1/5+1/5-1/7+...+1/11-1/13)`
`=1/2.(1/3-1/13)`
`=1/2 . 10/39`
`=5/39`
HH
1
22 tháng 7 2015
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\cdot\frac{4}{15}=\frac{2}{15}\)
HT
3
19 tháng 1 2018
ta có
(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)
=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)
19/36+17/90+1/12
=(19/36+1/12)+17/90
=7/12+17/90
=105/180+34/180
=139/180
19 tháng 1 2018
1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45
=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5
=1/1-1/3+1/3-1/2....+1/9-1/5
=1/1
LT
2
29 tháng 3 2016
=\(\frac{1}{3\cdot5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{11\cdot13}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{13}=\frac{5}{13}\)
HM
29 tháng 3 2016
B=1/15+1/35+1/63+1/99+1/143
B=1/3.5+1/5.7+1/7.9+1/9.11+11.13 (khoảng cách từ 3-5;5-7;7-9;9-11;11-13 la 2)
Suy ra B=1/2(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13) (Ta gop -1/5+1/5;-1/7+1/7;-1/9+1/9;-1/11+1/11 bang 0)
B=1/2(1/3-1/43)=1/2.40/129=20/129
DN
4
6 tháng 3 2017
A=1/3.5+1/5.7+1/7.9+1/9.11+1/11.13
A=1/2 [1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13]
A=1/2.10/39
A=5/39
ai tk mình
mình tk lại
TH
7
HH
15 tháng 6 2020
\(a)\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{22}{132}+\frac{11}{132}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{33}{132}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{34}{132}+\frac{1}{20}\)
\(=\frac{17}{66}+\frac{1}{20}\)
\(=\frac{203}{660}\)
15 tháng 6 2020
\(a,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\frac{3}{10}+\frac{1}{132}\)
\(=\frac{198}{660}+\frac{5}{660}\)
\(=\frac{203}{660}\)
6
2
H9
HT.Phong (9A5)
CTVHS
16 tháng 6 2023
\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{97}{98}\cdot\dfrac{98}{99}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot98}{2\cdot3\cdot4\cdot...\cdot99}\)
\(=\dfrac{1}{99}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
16 tháng 6 2023
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}=\dfrac{100}{2}=50\)
Ta có:
\(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{195}\)
\(\Rightarrow2A=\frac{2}{3}+\frac{2}{15}+\frac{2}{21}+...+\frac{2}{195}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{1}-\frac{1}{15}\)
\(=\frac{14}{15}\)
\(\Rightarrow2A=\frac{14}{15}\Rightarrow A=\frac{14}{15}\div2=\frac{7}{15}\)
Vậy A = 7/15
hình như đề sai ở ps cuối