Ta có :
\(\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2014}=\left(1-\dfrac{1}{2015}\right)+\left(1-\dfrac{1}{2016}\right)+\left(1+\dfrac{2}{2014}\right)\) \(=\left(1+1+1\right)-\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)\)
\(=3-\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)\)
Dễ thấy : \(\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)>0\) vì \(\dfrac{1}{2015}>\dfrac{1}{2016}\)
Do đó \(\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2014}>3\)
~ Học tốt ~
@Lâm Gia Bảo lập luận sai --> đáp số đúng là sao?
\(\dfrac{2014}{2015}=1-\dfrac{2014}{2015}\)
\(\dfrac{2015}{2016}=1-\dfrac{1}{2016}\)
\(\dfrac{2016}{2014}=1+\dfrac{2}{2014}\)
công lại
\(VT=3+\left(\dfrac{1}{2014}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)
dễ dàng nhận ra
\(\left\{{}\begin{matrix}\dfrac{1}{2014}>\dfrac{1}{2015}\\\dfrac{1}{2014}>\dfrac{1}{2016}\end{matrix}\right.\) \(\Rightarrow VT>3\)
