bài 9:cho :
C=\(\frac{2005}{2}+\frac{20025}{3}+\frac{2005}{4}+...+\frac{2005}{2005}\)
D=\(\frac{2006}{1}+\frac{2007}{2}+\frac{2008}{3}+...+\frac{4004}{2004}\)
tính C-D
K
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16 tháng 8 2015
=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8
8 tháng 5 2018
\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
23 tháng 2 2020
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .
HH
13 tháng 11 2020
\(C=\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}}\)
Đặt N = \(\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}\)
\(\Rightarrow N=\frac{1}{2006}+.....+\frac{2004}{3}+\frac{2005}{2}+\frac{2006}{1}\)
\(\Rightarrow N=\left(\frac{1}{2006}+1\right)+.....+\left(\frac{2004}{3}+1\right)+\left(\frac{2005}{2}+1\right)+1\)( Có 2005 nhóm )
\(=\frac{2007}{2006}+....+\frac{2007}{3}+\frac{2007}{2}+\frac{2007}{2007}\)
\(=2007\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2006}+\frac{1}{2007}\right)\)
Đặt M = \(\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}\)
\(=2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)\)
Thay N và M vào C , ta có :
\(C=\frac{N}{M}=\frac{2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)}{2007\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2007}\right)}=\frac{2006}{2007}\)
\(\Rightarrow C=\frac{2006}{2007}\)
Vậy : \(C=\frac{2006}{2007}\)
TH
2 tháng 4 2017
Đặt biểu thức là A ta có:
\(A=\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+...+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+...+\frac{1}{2006}}\)
\(\Rightarrow A=\frac{2006.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}\right)}{1+\left(1+\frac{2005}{2}\right)+\left(1+\frac{2004}{3}\right)+...+\left(1+\frac{1}{2006}\right)}\)
\(\Rightarrow A=\frac{2006.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)}{1+\frac{2007}{2}+\frac{2007}{3}+...+\frac{2007}{2006}}\)
\(\Rightarrow A=\frac{2006.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)}{2007.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}+\frac{1}{2007}\right)}\)
\(\Rightarrow A=\frac{2006}{2007}\)
HT
29 tháng 8 2015
Xét tử ta có:
\(2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{1}{2008}\)
= \(1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{1}{2008}\right)\)
= \(\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}\)
= \(2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\)
=> A = \(\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}\)
=> A = 2009
29 tháng 8 2015
A=\(\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...........+\left(1+\frac{2}{2008}\right)+\left(1+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2008}+\frac{1}{2009}}\)=\(\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+....+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\)
=\(\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\)
=2009
Vay A=2009
TG
23 tháng 10 2016
Gọi a là tử số, b là mẫu số của phân số A
a = \(\frac{2008}{1}\)+ \(\frac{2007}{2}\)+ \(\frac{2006}{3}\)+ ... + \(\frac{1}{2008}\)
Dãy số a có (2008 - 1) : 1 + 1 = 2008 số. Và a = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)
b = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{2009}\)
Dãy số b có (2009 - 2) : 1 + 1 = 2008 số. Và b = (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)
A = [ ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)] : [ (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)] = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) : (\(\frac{1}{2}\)+ \(\frac{1}{2009}\))
A = \(\frac{\text{2008 x2008 + 1}}{2008}\)x \(\frac{2x2009+2}{2x2009}\)
A = 2008