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2003/2004 + 2004/2005 + 2005/2003

= 1 - 1/2004 + 1 - 1/2005 + 1 + 1/2003 + 1/2003

=(1+1+1)-(1/2004 - 1/2003 + 1/2005 - 1/2003)

= 3 - (1/2004 - 1/2003 + 1/2005 - 1/2003)

Vì 1/2004 < 1/2003 ; 1/2005 < 1/2003

=>1/2004 - 1/2003 + 1/2005 - 1/2003 < 0

=> 3 - (...) > 3

Vậy. ...

K mình nha 

\(\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}=1-\frac{1}{2004}+1-\frac{1}{2005}+1+\frac{2}{2003}\)

\(=3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)\)

Do \(\frac{1}{2003}>\frac{1}{2004}>\frac{1}{2005}.\) nên \(\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>0\)

Vì vậy \(3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>3\) (đpcm)

\(A=\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}\)

\(=(1-\frac{1}{2004})+(1-\frac{1}{2005})+(1+\frac{2}{2003})\)

\(=3+(\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005})\)

Do\(\frac{1}{2003}\)>\(\frac{1}{2004}\)>\(\frac{1}{2005}\)

\(\Rightarrow\frac{1}{2003}+\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\)>\(0\)

\(\Rightarrow3+(\frac{1}{2003}-\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2005})\)>\(3\)

\(\Rightarrow A\)>\(3\)

GIUP MINH VOI NHA

A > B vì:

2005 > 2004

Bạn tự làm tiếp nhé

KO HIEU CACH GIAI

ko cần lm nhìn đề là tui biết a > b rùi

A>B