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\(\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}\)
\(\Rightarrow\frac{2008}{2006}>1\)
\(\frac{2006}{2007}< 1;\frac{2007}{2008}< 1\)
\(\Rightarrow\frac{2006}{2007}+\frac{2007}{2008}< 2\)
\(\Rightarrow\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}< 3\)
A =2006/2007+2007/2008+2008/2006
= \(\frac{2006}{2007}\)+ \(\frac{2007+1}{2008}\)+ \(\frac{2008}{2006+2}\)
= 1 - \(\frac{1}{2007}\)+ 1 - \(\frac{1}{2008}\)+ 1 + \(\frac{1}{2006}\)+ \(\frac{1}{2006}\)
= 3 + ( \(\frac{1}{2006}\)- \(\frac{1}{2007}\)) + ( \(\frac{1}{2006}\)- \(\frac{1}{2008}\))
vì \(\frac{1}{2006}\)> \(\frac{1}{2007}\), \(\frac{1}{2006}\)> \(\frac{1}{2008}\)nên A > 3
2.006/2.007 + 2.007/2.008 < 2006 + 2.007/2.007 + 2.008
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\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}\)
\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{2}{2006}\)
\(A=\left(1+1+1\right)+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)
\(A=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)
Ta thấy : \(\frac{1}{2006}-\frac{1}{2007}>0\); \(\frac{1}{2006}-\frac{1}{2008}>0\)\(\Rightarrow A>3\)
2005/2006 + 2006/2007 + 2007/2008 + 2008/2005
= 4,000001491
k minh di xin day
minh ko bietcach giai tra loi giup minh di ban minh can gap
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}=1-\frac{1}{2007}+1-\frac{1}{2008}+1+\frac{2}{2006}=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)
Vì 2006<2007, 2006<2008 nên \(\frac{1}{2006}>\frac{1}{2007};\frac{1}{2006}>\frac{1}{2008}=>\frac{1}{2006}-\frac{1}{2007}>0,\frac{1}{2006}-\frac{1}{2008}>0\)
=> \(A=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)>3=>A>3\)
bạn ơi cái dấu bằng to và dấu lớn to là dấu suy ra ak
Ta có: 3 = 1 + 1 + 1
Ta có: 2006/2007 < 1 ; 2007/2008 < 1 ; 2008/2009 < 1
Nên 2006/2007 + 2007/2008+ 2008/2009 < (1+1+1=3)
Ta có: \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}\)
\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{2}{2006}\)
\(A=3+\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2008}>3\)
Vậy A > 3
A=20062007+20072008+20082009=1−12007+1−12008+1−12009�=20062007+20072008+20082009=1−12007+1−12008+1−12009
=3−12007−12008−12009
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