1)\(P=\left|a-\dfrac{1}{2014}\right|+\left|a-\dfrac{1}{2016}\right|=\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\)
Cái này tự tính được nhé
2) \(\dfrac{6}{x+1}.\dfrac{x-1}{3}\in Z\Leftrightarrow\dfrac{6\left(x-1\right)}{3\left(x+1\right)}\in Z\)
\(\Rightarrow6x-6⋮3x+3\)
\(\Rightarrow6x+6-12⋮3x+3\)
\(\Rightarrow12⋮3x+3\)
Ok:>
Câu 1:
Thay \(a=\dfrac{1}{2015}\) vào biểu thức \(P=\left|a-\dfrac{1}{2014}\right|+\left|a-\dfrac{1}{2016}\right|\) ta được:
\(\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\)
\(=\left|\dfrac{2014}{4058210}-\dfrac{2015}{4058210}\right|+\left|\dfrac{2016}{4062240}-\dfrac{2015}{4062240}\right|\)
\(=\left|\dfrac{2014-2015}{4058210}\right|+\left|\dfrac{2016-2015}{4062240}\right|\)
\(=\left|-\dfrac{1}{4058210}\right|+\left|\dfrac{1}{4062240}\right|\)
\(=\dfrac{1}{4058210}+\dfrac{1}{4062240}\)
\(=\dfrac{1008}{4090695680}+\dfrac{1007}{4090695680}\)
\(=\dfrac{1008+1007}{4090695680}\)
\(=\dfrac{2015}{4090695680}\)
\(=\dfrac{2015}{4090695680}\)
\(=\dfrac{1}{2030112}\)
Vậy giá trị của biểu thức P tại \(a=\dfrac{1}{2015}\) là \(\dfrac{1}{2030112}\)
