vì |x+2017|\(\ge\)0
=> |x+2017|+2018\(\ge\)2018
|x+2017|+2019\(\ge\)2019
=> GTNN của \(\dfrac{\left|x+2017\right|+2018}{\left|x+2017\right|+2019}\)=\(\dfrac{2018}{2019}\)
Đặt \(t=\left|x+2017\right|\ge0\)
Đặt biểu thức là T, ta có:
\(T=\dfrac{t+2018}{t+2019}=\dfrac{t+2019-1}{t+2019}=1-\dfrac{1}{t+2019}\)
Ta có: \(t\ge0\Rightarrow t+2019\ge2019\)
\(\Rightarrow\dfrac{1}{t+2019}\le\dfrac{1}{2019}\)
\(\Rightarrow-\dfrac{1}{t+2019}\ge-\dfrac{1}{2019}\)
\(\Rightarrow T\ge1-\dfrac{1}{2019}=\dfrac{2008}{2009}\)
GTNN của T là \(\dfrac{2008}{2009}\) khi \(t=0\Leftrightarrow\left|x+2017\right|=0\Leftrightarrow x=-2017\)
