C= (1+2/3).(1+2/5).(1+2/7).......(1+2/2009).(1+2/2011)
Những câu hỏi liên quan
`C=(1+2/3)(1+2/5)(1+2/7)...(1+2/2009)(1+2/2011)`
`#P.Vy`
`C=(3/3+2/3)(5/5+2/5)(7/7+2/7)...(2009/2009+2/2009)(2011/2011+2/2011)`
`C=5/3xx7/5xx9/7xx...xx2011/2009xx2013/2011`
\(C=\dfrac{5.7.9\times...\times2011.2013}{3.5.7\times...\times2009.2011}\)
`C=671`
Đúng 4
Bình luận (0)
https://olm.vn/hoi-dap/detail/105210184248.html
Đúng 3
Bình luận (0)
C=(1+2/3). (1+2/5).(1+2/7)... (1+2/2009).(1+2/2011) = ?
C=5/3×7/5×9/7×...2011/2009×2013/2011
C=5×7×9×...×2011×2013/3×5×7×...×2009×2011
C=2013/3
C=674
Nếu sai bảo mình nhé !
🤓🤓🤓🤓🤓🤓🤓🤓🤓🤓🤓🤓😸😸😸😸😸😸😸😸😸😸🐒🐒🐒🐒🐒
Đúng 0
Bình luận (0)
tính các dãy số sau:
-1-2+3+4-5-6+7+8-.....-2009-2010+2011
-1-2+3-4-5+6+....-2009+2010-2011
C=\((1+\frac{2}{3})\times(1+\frac{2}{5})\times(1+\frac{2}{7})\times...\times(1+(\frac{2}{2009})\times(1+\frac{2}{2011})\)
\(C=\left(1+\frac{2}{3}\right)\cdot\left(1+\frac{2}{5}\right)\cdot\left(1+\frac{2}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1+\frac{2}{2009}\right)\cdot\left(1+\frac{2}{2011}\right)\)
\(C=\frac{5}{3}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\cdot\cdot\cdot\cdot\frac{2011}{2009}\cdot\frac{2013}{2011}\)
\(C=\frac{5\cdot7\cdot9\cdot\cdot\cdot\cdot\cdot2011\cdot2013}{3\cdot5\cdot7\cdot\cdot\cdot\cdot\cdot2009\cdot2011}\)
\(C=\frac{2013}{3}\)
Đúng 0
Bình luận (0)
đề bài của bài là tính
ai nhanh mk k cho
Đúng 0
Bình luận (0)
\(C=\left(1+\frac{2}{3}\right).\left(1+\frac{2}{5}\right).....\left(1+\frac{2}{2009}\right).\left(1+\frac{2}{2011}\right)\)
\(C=1\left(\frac{2}{3}+\frac{2}{5}+...+\frac{2}{2009}+\frac{2}{2011}\right)\)
\(\frac{1}{2}C=\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2009}+\frac{1}{2011}\)
\(\frac{1}{2}C=\frac{1}{3+5+...+2011}\)
\(\frac{1}{2}C=\frac{1}{1012035}\)
\(\Rightarrow C=\frac{1}{1012035}:\frac{1}{2}=\frac{2}{1012035}\)
Đúng 0
Bình luận (0)
Tính tổng các đại số sau
a) -1-2-3-4-5-............-2009-2010
b) 1-3+5-7+...... ..+2005-2007+2009-2011
c) 1-2-3+4+5-6-7+...............+1997-1998-1999+2000+2001
a) -1 - 2 - 3 - 4 - 5 -.............- 2009 - 2010
SCSH: ( 2010 - 1 ) : 1 + 1 = 2010
tỔNG: ( 2010 + 1 ) . 2010 : 2 = 2021055
b) 1 - 3 + 5 - 7 +...............+ 2005 - 2007 + 2009 - 2011
SCSH: ( 2011 - 1 ) : 2 + 1 = 1006
tỔNG: ( 2011 + 1 ) . 1006 : 2 = 1012036
c) 1 - 2 - 3 + 4 + 5 - 6 - 7 +..........................+ 1997 - 1998 - 1999 + 2000 + 2001
SCSH: ( 2001 - 1 ) : 1 + 1 = 2001
tỔNG: ( 2001 + 1 ) . 2001 : 2 = 2003001
Hk tốt,
k nhé
Đúng 0
Bình luận (0)
1. Tính
a) 6/5 + 6/35 - 6/125 - 6/2009 - 6/2011
7/5 + 7/35 + 7/125 + 7/2009 + 7/2011
b)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{1998^2}\right)\)
c) (1+2/3)x(1+2/5)x(1+2/7)...(1+2/2009)x(1+2/2011)
các bn giúp mk nha
Đặt \(C=\left(1+\frac{2}{3}\right)\left(1+\frac{2}{5}\right)\left(1+\frac{2}{7}\right).....\left(1+\frac{2}{2009}\right)\left(1+\frac{2}{2011}\right)\) ta có :
\(C=\left(\frac{3+2}{3}\right)\left(\frac{5+2}{3+2}\right)\left(\frac{7+2}{5+2}\right).....\left(\frac{2009+2}{2007+2}\right)\left(\frac{2011+2}{2009+2}\right)\)
\(C=\frac{\left(3+2\right)\left(5+2\right)\left(7+2\right).....\left(2009+2\right)\left(2011+2\right)}{3\left(3+2\right)\left(5+2\right).....\left(2007+2\right)\left(2009+2\right)}\)
\(C=\frac{2011+2}{3}\)
\(C=\frac{2013}{3}\)
\(C=671\)
Vậy \(C=671\)
Chúc bạn học tốt ~
Đúng 0
Bình luận (0)
\(\left(1+\frac{2}{3}\right)\cdot\left(1+\frac{2}{5}\right)\cdot\left(1+\frac{2}{7}\right)...\left(1+\frac{2}{2009}\right)\cdot\left(1+\frac{2}{2011}\right)\)
\(=\frac{5}{3}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\cdot\cdot\frac{2011}{2009}\cdot\frac{2013}{2011}\)
\(=\frac{2013}{3}=671\)
Đúng 0
Bình luận (0)
ai đó giúp mk với mk xin chân thành cảm ơn! a=(2010+2010/2+2009/3+2008/4+...+1/2011/ 1/2+1/3+...+1/2011) / (1/2+1/3+1/4+1/5+...+1/2009+1/2010+1/2011)
\(C=\left(1+\frac{2}{3}\right).\left(1+\frac{2}{5}\right).\left(1+\frac{2}{7}\right).....\left(1+\frac{2}{2009}\right)+\left(1+\frac{2}{2011}\right)\)
C=(1+2/3).(1+2/5).(1+2/7)......(1+2/2009).(1+2/2011)
C=5/3.7/5.9/7......2011/2009.2013/2011
C=5.7.9.....2013/3.5.7.....2009.2011
C=2013/3
Đúng 0
Bình luận (0)