B
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NN
2
15 tháng 6 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\left(\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2010}{2011}=\frac{1005}{2011}\)
JW
15 tháng 6 2016
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)=\frac{1}{2}.\left(1-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2010}{2011}=\frac{1005}{2011}\)
PM
6
11 tháng 5 2019
A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2011.2013
A = 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2011.2013)
A = 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2011 - 1/2013)
A = 1/2.(1 - 1/2013)
A = 1/2.2012/2013
A = 1006/2013
11 tháng 5 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(2A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{2011}-\frac{1}{2011}\right)-\frac{1}{2013}\)
\(2A=1-\frac{1}{2013}\)
\(2A=\frac{2012}{2013}\)
\(A=\frac{2012}{2013}:2\)
\(A=\frac{1006}{2013}\)
~ Hok tốt ~
ZV
2
15 tháng 6 2016
1/.3 + 1/3.5 + 1/5.7 + ... + 1/2009.2011
= 1/2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2009.2011)
= 1/2 . (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2009 - 1/2011)
= 1/2 . (1 - 1/2011)
= 1/2 . 2010/2011
= 1005/2011
KC
15 tháng 6 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)
\(=\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}x\left(1-\frac{1}{2011}\right)\)
\(=\frac{1005}{2011}\)
VT
1
16 tháng 4 2019
a, 1 + 2 + 3 + ... + x = 120
=> (x+1)x/2 = 120
=>x(x +1)=120.2=240
=>15.16 = 240
=>x=15
Vậy x=15
Phần b làm tương tự
c, x - ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/53.55) = 3/5
=> x = 3/5 + ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/53.55)
=> x = 3/5 + ( 1-1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/53 - 1/55 )
=> x = 3/5 + ( 1- 1/55 )
=> x = 3/5 + 54/55
=> x = 87/55
Vậy x = 87/55
28 tháng 4 2019
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\frac{2012}{2013}\)
\(A=\frac{1006}{2013}\)
28 tháng 4 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\frac{2012}{2013}\)
\(A=\frac{1006}{2013}\)
KN
28 tháng 4 2019
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(\Rightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\)
\(\Rightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(\Rightarrow2S=1-\frac{1}{2013}\)
\(\Rightarrow2S=\frac{2012}{2013}\)
\(\Rightarrow S=\frac{2012}{2013}\div2\)
\(\Rightarrow S=\frac{1006}{2013}\)
BD
28 tháng 4 2019
\(2S=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2011\cdot2013}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(2S=1-\frac{1}{2013}\)
\(2S=\frac{2012}{2013}\)
\(S=\frac{2012}{2013}\div2=\frac{1006}{2013}\)
#Louis
Gọi \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(\Rightarrow2A=1-\frac{1}{2013}\)
\(\Rightarrow2A=\frac{2012}{2013}\)
\(\Rightarrow A=\frac{1006}{2013}\)