Ta có :\(\frac{2012}{\sqrt{2013}}+\frac{2013}{\sqrt{2012}}=\frac{2013-1}{\sqrt{2013}}+\frac{2012+1}{\sqrt{2012}}\)

=>\(\frac{2013}{\sqrt{2013}}-\frac{1}{\sqrt{2013}}+\frac{2012}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}\)

=>\(\sqrt{2013}-\frac{1}{\sqrt{2013}}+\sqrt{2012}+\frac{1}{\sqrt{2012}}\)

Mà \(\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}>0\)

Vậy \(\sqrt{2012}+\sqrt{2013}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}>\sqrt{2012}+\sqrt{2013}\)

Hay \(\frac{2012}{\sqrt{2013}}+\frac{2013}{\sqrt{2012}}>\sqrt{2012}+\sqrt{2013}\)

= \(\frac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\frac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{1}{\sqrt{2012}.\sqrt{2013}\left(\sqrt{2013}+\sqrt{2012}\right)}\)

= \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}+...+\frac{\left(\sqrt{2013}-\sqrt{2012}\right)\left(\sqrt{2013}+\sqrt{2012}\right)}{\sqrt{2012}\sqrt{2013}\left(\sqrt{2012}+\sqrt{2013}\right)}\)

= \(\frac{\sqrt{2}-1}{\sqrt{2}}+...+\frac{\sqrt{2013}-\sqrt{2012}}{\sqrt{2012}\sqrt{2013}}\)

= \(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\)

= \(\frac{\sqrt{2013}-1}{\sqrt{2013}}=\frac{2013-\sqrt{2013}}{2013}\)

đặt \(A=\frac{2012}{\sqrt{2013}}+\frac{2013}{\sqrt{2012}};B=\sqrt{2012}+\sqrt{2013}\)

ta có:\(A=\frac{2013-1}{\sqrt{2013}}+\frac{2012+1}{\sqrt{2012}}=\sqrt{2013}-\frac{1}{\sqrt{2013}}+\sqrt{2012}+\frac{1}{\sqrt{2012}}\)

\(\Rightarrow A=\left(\sqrt{2013}+\sqrt{2012}\right)+\left(\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)>\sqrt{2012}+\sqrt{2013}=B\)

vậy A>B(đpcm)
 

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