5/1.3+5/3.5+5/5.7+...+5/91.93+5/93.95
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Những câu hỏi liên quan
HA
0
HM
2
YS
5 tháng 5 2016
= 5- 5/3 + 5/3 - 5/5 + ... + 5/93 - 5/95
= 5 -5/95
= 90/95
Đ/s: 90/95
CX
5 tháng 5 2016
5/1.3 + 5/3.5 +5/5.7+....+5/91.93 + 5/93.95
= 5/2 . (1-1/3+1/3-1/5+1/5-1/7+....+1/91-1/93+1/93-1/95)
= 5/2 . (1-1/95)
= 5/2 . 94/95
= 47/19
Nhớ k nha
5 tháng 5 2016
Ta có:
\(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{91.93}+\frac{5}{93.95}=5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}+\frac{1}{93.95}\right)=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{91.93}+\frac{2}{93.95}\right)\)
\(\Rightarrow A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{93}+\frac{1}{93}-\frac{1}{95}\right)=\frac{5}{2}\left(1-\frac{1}{95}\right)=\frac{5}{2}.\frac{94}{95}=\frac{47}{19}\)
Vậy \(A=\frac{47}{19}\)
MC
5 tháng 5 2016
\(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{93.95}\)
\(A=5\cdot\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{95}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{95}\right)=\frac{5}{2}\cdot\frac{94}{95}=\frac{47}{19}\)
5 tháng 5 2016
\(\frac{2}{5}A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{93.95}\)
\(\frac{2}{5}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{93}-\frac{1}{95}\)
\(\frac{2}{5}A=1-\frac{1}{95}=\frac{94}{95}\)
\(A=\frac{47}{19}\)
CX
5 tháng 5 2016
A= 5/1.3 +5/3.5 +5/5.7+....+5/91.93 + 5/93.95
A= 5/2 . (1-1/3+1/3-1/5+1/5-1/7+...+1/91-1/93+1/93-1/95)
A=5/2. (1-1/95)
A= 5/2 . 94/95
A=47/19
Nhớ k nha
NT
9
28 tháng 4 2015
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
HD
1
TT
14 tháng 4 2016
a.2/1.3+2/3.5+2/5.7+................+2/99.101
1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
1-1/101
100/101
b.5/1.3+5/3.5+5/5.7+............+5/99.101
5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2
5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)
5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)
5/2(1-1/101)
5/2.100/101
250/101
3 tháng 5 2015
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2
=(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2
=(1/2-1/2010).2
=1004/1005
6 tháng 3 2017
A=5/1.3+5/3.5+5/5.7+...+5/99.101
A=5.1/2 [1/1-1/3+1/3-1/5+1/5-1/7+1/7+...+1/99-1/101 ]
A=5.1/2.100/101
A=250/101
k nha
TG
2
9 tháng 5 2023
\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{201\cdot203}\)
= \(\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{201\cdot203}\right)\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)
= \(\dfrac{5}{2}\cdot\dfrac{202}{203}=\dfrac{505}{203}\)
TA
9 tháng 5 2023
Ta có :
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{201.203}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{201.203}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)
\(=\dfrac{5}{2}.\dfrac{202}{203}\)
\(=\dfrac{505}{203}\)
Đặt A= 5/1.3 + 5/3.5 + 5/5.7 + ... + 5/91.93 + 5/93.95
=> 2/5 A = 2/5 . (5/1.3 + 5/3.5 + 5/5.7 + ... + 5/91.93 + 5/93.95)
<=> 2/5A = 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/91.93 + 2/93.95
<=> 2/5A = 1-1/3 +1/3-1/5+1/5-1/7+...+1/91-1/93+1/93-1/95
<=> 2/5A = 1-1/95
<=> 2/5A = 94/95
=> A = 94/95 : 2/5
<=> A = 47/19