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\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.....+\frac{1}{80}\)
\(=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+......+\frac{1}{80}\right)\)
\(>\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+.....+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\right)\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)
Giải:
Ta có:
A=20092008+1/20092009+1
2009A=20092009+2009/20092009+1
2009A=20092009+1+2008/20092009+1
2009A=20092009+1/20092009+1 + 2008/20092009+1
2009A=1+2008/20092009+1
Tương tự:
B=20092009+1/20092010+1
2009B=1+2008/20092010+1
Vì 2008/20092009+1 > 2008/20092010+1 nên 2009A>2009B
⇒A>B
