Chứng minh rằng: A= 2010/2011+2011/2012+2012/2010 > 3
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Có : \(2009+2010>\dfrac{2009}{2010}\) ; \(2011+2012>\dfrac{2011}{2012}\)
\(\dfrac{2011}{2010}>1\) ; \(\dfrac{2010}{2011}< 1\) \(\Rightarrow\dfrac{2011}{2010}>\dfrac{2010}{2011}\)
Ta có : \(2009+2010+\dfrac{2011}{2010}+2011+2012>\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2011}{2012}\)
\(\Leftrightarrow B>A\)
Hay \(A< B\)
Ta có: \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
Nên \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)\(\Rightarrow A>B\)
So sánh: \(\frac{2010}{2011}+\frac{2011}{2012}\) với \(\frac{2010+2011}{2011+2012}\)
\(P=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}=\frac{2010}{8144863716}+\frac{2011}{8144863716}+\frac{2012}{8144863716}\)
\(=\frac{6033}{8144863716}=\frac{1}{1350052}\)
\(Q=2010+2011+\frac{2012}{2011}+2012+2013\)
\(=2010+2011+2012+2013+\frac{2012}{2011}\)
\(=8046+\frac{2012}{2011}=\frac{8046}{1}+\frac{2012}{2011}\)
\(=\frac{16180506}{2011}+\frac{2012}{2011}=\frac{16182518}{2011}\)
Lời giải:
$A=1-\frac{1}{2011}+1-\frac{1}{2012}+1+\frac{2}{2010}$
$=3+(\frac{1}{2010}-\frac{1}{2011})+(\frac{1}{2010}-\frac{1}{2012})$
$> 3+0+0+0=3$
Ta có đpcm.