so sánh tổng A với 3
A= 2005/2006+2006/2007+2007/2005
giúp tui nhé tui cần
K
Khách
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Những câu hỏi liên quan
NX
0
NQ
1
16 tháng 5 2015
Ta thấy:
2005/2006 = 1 - 1/2006
2006/2007 = 1 - 1/2007
2007/2008 = 1 - 1/2008
2008/2005 = 1 + 3/2005
Mà: 1/2005 > 1/2006 > 1/2007 > 1/2008
=> 3/2005 - 1/2006 - 1/2007 - 1/2008 > 0
=> 2005/2006 + 2006/2007 + 2007/2008 + 2008/2005 > 4
PK
1
NT
0
15 tháng 5 2017
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>\frac{2001}{2001}+\frac{2002}{2002}+\frac{2003}{2003}+\frac{2004}{2004}+\frac{2005}{2005}+\frac{2006}{2006}+\frac{2007}{2007}+\frac{2008}{2008}\)
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>1+1+1+1+1+1+1+1\)\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
\(A>8\)
20023
Ta có: \(A=\dfrac{2005}{2006}+\dfrac{2006}{2007}+\dfrac{2007}{2005}=\dfrac{2006-1}{2006}+\dfrac{2007-1}{2007}+\dfrac{2005}{2005}+\dfrac{1}{2005}+\dfrac{1}{2005}\)\(=1+1+1+\left(\dfrac{1}{2005}-\dfrac{1}{2006}+\dfrac{1}{2005}-\dfrac{1}{2007}\right)\)
\(=3+\left(\dfrac{1}{2005}-\dfrac{1}{2006}+\dfrac{1}{2005}-\dfrac{1}{2007}\right)\)
Ta thấy: \(\dfrac{1}{2005}>\dfrac{1}{2006};\dfrac{1}{2005}>\dfrac{1}{2007}\) \(\Rightarrow\dfrac{1}{2005}-\dfrac{1}{2006}+\dfrac{1}{2005}-\dfrac{1}{2007}>0\)
\(\Rightarrow A>3\)