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Ta có: \(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+4}{2011}+\frac{x+5}{2010}+\frac{x+6}{2009}\)
\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1=\frac{x+4}{2011}+1+\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
\(\Rightarrow\frac{2015+x}{2014}+\frac{2015+x}{2013}+\frac{2015+x}{2012}=\frac{2015+x}{2011}+\frac{2015+x}{2010}+\frac{2015+x}{2009}\)
\(\Rightarrow\left(2015+x\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> 2015 + x = 0
=> x = -2015
\(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2013}{1}+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}-2012}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2013}{1}-1\right)+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4024}{2012}-1\right)}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2012}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
Ủng hộ mk nha ^_-
\(\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{2011}{2012}.\frac{2013}{2014}\right)^2\)
\(=\left(\frac{1}{2}\right)^2\left(\frac{3}{4}\right)^2...\left(\frac{2013}{2014}\right)^2\)
\(< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{2013}{2014}.\frac{2014}{2015}=\frac{1}{2015}\)
\(\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{2011}{2012}.\frac{2013}{2014}\right)^2\)
\(=\left(\frac{1}{2}\right)^2\left(\frac{3}{4}\right)^2...\left(\frac{2013}{2014}\right)^2\)
\(>\frac{1}{2}.\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2012}{2013}.\frac{2013}{2014}=\frac{1}{2}.\frac{1}{2014}=\frac{1}{4028}\)
Suy ra đpcm.