Chứng minh rằng:
\(2005^{2007}+2007^{2005}⋮2006\)
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2005/2006 + 2006/2007 + 2007/2008 + 2008/2005
= 4,000001491
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minh ko bietcach giai tra loi giup minh di ban minh can gap
`2007/2009×2002/2005×2009/2006×2005/2007×2006/2002`
`=(2007xx2002xx2009xx2005xx2006)/(2009xx2005xx2006xx2007xx2002)`
`=(2007xx2002xx2009xx2005xx2006)/(2007xx2002xx2009xx2005xx2006)`
`=1`
\(\dfrac{2007}{2009}.\dfrac{2002}{2005}.\dfrac{2009}{2006}.\dfrac{2005}{2007}.\dfrac{2006}{2002}\\ =\left(\dfrac{2007}{2009}.\dfrac{2009}{2006}\right).\left(\dfrac{2006}{2002}.\dfrac{2002}{2005}\right).\dfrac{2005}{2007}\\ =\dfrac{2007}{2006}.\dfrac{2006}{2005}.\dfrac{2005}{2007}=1\)
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: \(\dfrac{1}{2005}>\dfrac{1}{2006};\dfrac{1}{2005}>\dfrac{1}{2007}\) \(\Rightarrow\dfrac{1}{2005}-\dfrac{1}{2006}+\dfrac{1}{2005}-\dfrac{1}{2007}>0\)
\(\Rightarrow A>3\)
Ta có: 20052007 + 20072005 = (20052007 + 12007 ) + (20072005 - 12005 )
Vì \(2005^{2007}+1^{2007}\)luôn chia hết cho \(2005+1=2006\left(1\right)\)
\(2007^{2005}-1^{2005}\)luôn chia hết cho \(2007-1=2006\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\left(2005^{2007}+1^{2007}\right)+\left(2007^{2005}-1^{2005}\right)⋮2006\)
\(\Rightarrow2005^{2007}+2007^{2005}⋮2006\)
Vậy \(2005^{2007}+2007^{2005}⋮2006\)