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Vì 2006/2007 ; 2007/2008 ; 2008/2009 ; 2009/2010 đều bé hơn 1 nên:
2006/2007 + 2007/2008 + 2008/2009 + 2009/2010 < 1 + 1 + 1 + 1 = 4.
Vậy ...
Ta có : \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)
\(=\frac{2007-1}{2007}+\frac{2008-1}{2008}+\frac{2009-1}{2009}+\frac{2006+3}{2006}\)
\(=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)
\(=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)
\(< 4-\left(\frac{1}{2009}+\frac{1}{2009}+\frac{1}{2009}-\frac{3}{2009}\right)\)
\(=4\)
=> A < 4
Vậy A < 4
Bài 1:
Ta có: 200920=(20092)10=403608110 ; 2009200910=2009200910
Vì 403608110< 2009200910 => 200920< 2009200910
Bài 1:
Ta có:\(2009^{20}\)=\(2009^{10}\).\(2009^{10}\)
\(20092009^{10}\)=(\(\left(2009.10001\right)^{10}=2009^{10}.10001^{10}\)
Vì 2009<10001\(\Rightarrow2009^{20}< 20092009^{10}\)
Ta có A= (1 -1/2007) +(1-1/2008)+(1-1/2009)+(1+3/2006)= 4-(1/2007+1/2008+1/2009-3/2006) <4
Đề của bạn sai rồi: Phải là B = \(\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\) chứ ?!
ta có: \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)
A = \(1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)
A= \(4\)\(+\frac{3}{2006}-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
Do 1/2007 < 1/2006 ; 1/2008<1/2006 ; 1/2009<1/2006=> 1/2007 + 1/2008 + 1/2009 < 1/2006 + 1/2006 + 1/2006
Mà 1/2006 + 1/2006 + 1/2006 = 3/2006
=> 3/2006 -( 1/2007 + 1/2008 + 1/2009) > 0
=> \(4+\frac{3}{2006}-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)>4\)
=> A > 4
Ta có:\(\frac{2006}{2007}< 1\)
\(\frac{2007}{2008}< 1\)
\(\frac{2008}{2009}< 1\)
\(\frac{2009}{2006}>1\)\(\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}< 4\)