\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)
Violympic toán 7
Các câu hỏi tương tự
\(\dfrac{2008}{1}\)+\(\dfrac{2007}{2}\)+\(\dfrac{2006}{3}\)+......+\(\dfrac{2}{2007}\)+\(\dfrac{1}{2008}\)
Help me
A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}\)
B = \(\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
Tính \(\dfrac{B}{A}\)
Tìm x,biết \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
Cho A=\(\dfrac{2006}{2007}+\dfrac{2007}{2008}+\dfrac{2008}{2009}+\dfrac{2009}{2006}\) .Hãy so sánh số đó với 4
Tìm x biết: \(\dfrac{x-1}{2009}\)+\(\dfrac{x-2}{2008}\)=\(\dfrac{x-3}{2007}\)+\(\dfrac{x-4}{2006}\)
Bài 1. a, Cho A left(dfrac{1}{2}-1right)left(dfrac{1}{3}-1right).....left(dfrac{1}{10}-1right)
So sánh A với dfrac{-1}{9}
Bài 2. Cho A left(dfrac{1}{2}-1right)left(dfrac{1}{3}-1right)....left(dfrac{1}{2008}-1right)left(dfrac{1}{2009}-1right)
B left(-1dfrac{1}{2}right)left(-1dfrac{1}{3}right)....left(-1dfrac{1}{2007}right)left(-1dfrac{1}{2008}right)
Tính A . B ?
Đọc tiếp
Bài 1. a, Cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\)
So sánh A với \(\dfrac{-1}{9}\)
Bài 2. Cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)....\left(\dfrac{1}{2008}-1\right)\left(\dfrac{1}{2009}-1\right)\)
B = \(\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)....\left(-1\dfrac{1}{2007}\right)\left(-1\dfrac{1}{2008}\right)\)
Tính A . B ?
tính hợp lý :
a, \(2008.\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009.\left(\dfrac{1}{2007}-2\right)\)
b, \(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3.4^5}\)
So sánh A và B, biết \(A=\dfrac{10^{2006}+1}{10^{2007}+1};B=\dfrac{10^{2007}+1}{10^{2008}+1}\)
Tính hợp lý:
a, \(2008\cdot\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\cdot\left(\dfrac{1}{2007}-2\right)\)
b,\(\dfrac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}\)