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Bài nãy sai rồi, cho mình làm lại nha:
\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}\)
\(=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{1}{2011}\)
Vì: \(\frac{1}{2011}>\frac{1}{2012}>\frac{1}{2013}\Rightarrow\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2012}+\frac{1}{2012}>0\)
\(\Rightarrow\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}>3\)
Nên \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)
b,Ta có
\(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
\(\Rightarrow P>Q\)
\(A=\frac{-10}{20}+\frac{-10}{30}+\frac{-10}{42}+\frac{-10}{56}+\frac{-10}{72}+\frac{-10}{90}+\frac{-10}{110}\)
\(=-10\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)\)
\(=-10\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=-10\left(\frac{1}{4}-\frac{1}{11}\right)\)
\(=\frac{-35}{22}\)
S= \(\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+2}{2011}\)
= 3 + \(\frac{2}{2011}-\frac{1}{2012}-\frac{1}{2013}\)
có \(\frac{1}{2011}>\frac{1}{2012}\)và \(\frac{1}{2011}>\frac{1}{2013}\)
\(\Rightarrow S>3\)
Bài giải :
Theo đề bài ra ta có : n. (n - 1) : 2 = 435
=> n. (n - 1) = 435 . 2 = 870
=> n.(n-1) = 30. 29
Vậy n = 30.
\(\frac{2010}{2011}\)> \(\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}\)> \(\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}\)> \(\frac{2012}{2011+2012+2013}\)
=> \(\frac{2010}{2011}\)+ \(\frac{2011}{2012}\)+ \(\frac{2012}{2013}\)> \(\frac{2010+2011+2012}{2011+2012+2013}\)
=> P > Q