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\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}=1-\frac{1}{2007}+1-\frac{1}{2008}+1+\frac{2}{2006}\)
\(=3-\frac{1}{2007}-\frac{1}{2008}+\frac{2}{2006}=3+\left(\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2008}\right)\)
Vì \(\frac{1}{2006}>\frac{1}{2007};\frac{1}{2006}>\frac{1}{2008}\Rightarrow\frac{1}{2006}-\frac{1}{2007}>0;\frac{1}{2006}-\frac{1}{2008}>0\)
Do đó \(\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2008}>0\)
\(\Rightarrow3+\left(\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2008}\right)>3\)
Vậy \(A>3\)
mỗi số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15
ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
hay tong tren be hon 15
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}\)
\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{2}{2006}\)
\(A=\left(1+1+1\right)+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)
\(A=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)
Ta thấy : \(\frac{1}{2006}-\frac{1}{2007}>0\); \(\frac{1}{2006}-\frac{1}{2008}>0\)\(\Rightarrow A>3\)
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
\(\frac{2006}{2007}< \frac{2007}{2007}=1\)
\(\frac{2007}{2008}< \frac{2008}{2008}=1\)
\(\frac{2008}{2009}< \frac{2009}{2009}=1\)
\(\Rightarrow a=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}< 1+1+1=3\)
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}\)
\(A=\left(1-\frac{1}{2007}\right)+\left(1-\frac{1}{2008}\right)+\left(1-\frac{1}{2009}\right)\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
\(A=3-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)< 3\)
\(A=\frac{2006+2007}{2006.2007}=\frac{2006}{2006.2007}+\frac{2007}{2006.2007}=\frac{1}{2007}+\frac{1}{2006}\)
\(B=\frac{2007+2008}{2007.2008}=\frac{2007}{2007.2008}+\frac{2008}{2007.2008}=\frac{1}{2008}+\frac{1}{2007}\)
Vì \(\frac{1}{2007}+\frac{1}{2006}>\frac{1}{2008}+\frac{1}{2007}\)
=> \(A>B\)
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{2}{2006}=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)\(>3+\left(\frac{1}{2007}-\frac{1}{2007}\right)+\left(\frac{1}{2008}-\frac{1}{2008}\right)=3=>A>3\)
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