Thay a=\(\dfrac{1}{2015}\) vào P, ta có:
\(P=\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\) (*)
Vì \(\left\{{}\begin{matrix}\dfrac{1}{2015}< \dfrac{1}{2014}\\\dfrac{1}{2015}>\dfrac{1}{2016}\end{matrix}\right.\) nên (*) \(\Rightarrow P=\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}=\dfrac{1}{2030112}\)
Vậy ...
\(P=\left|a-\dfrac{1}{2014}\right|+\left|a-\dfrac{1}{2016}\right|\\ =\left|\dfrac{1}{2014}-a\right|+\left|a-\dfrac{1}{2016}\right|\\ =\left|\dfrac{1}{2014}-\dfrac{1}{2015}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\\ =\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\\ =\dfrac{1}{2014}-\dfrac{1}{2016}\\ =\dfrac{2}{2014\cdot2016}\\ =\dfrac{1}{2030112}\)