Ngoài cách trên , mik xin trình bày cách 2 ạ
ĐKXĐ : x khác 0
\(A=\dfrac{x^2+2x+2018}{x^2}=1+\dfrac{2}{x}+\dfrac{2018}{x^2}\)
Đặt \(\dfrac{1}{x}=a\) , ta có :
\(A=1+2a+2018a^2\)
\(=2018\left(a^2+2a.\dfrac{1}{2018}+\dfrac{1}{2018^2}\right)+\dfrac{2017}{2018}\)
\(=2018\left(a+\dfrac{1}{2018}\right)^2+\dfrac{2017}{2018}\ge\dfrac{2017}{2018}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow a=-\dfrac{1}{2018}\Leftrightarrow\dfrac{1}{x}=-\dfrac{1}{2018}\Leftrightarrow x=-2018\)
Vậy ...
\(2018A=\dfrac{2018x^2+2.2018.x+2018^2}{x^2}=\dfrac{2017x^2}{x^2}+\dfrac{x^2+2.2018+2018^2}{x^2}=2017+\dfrac{\left(x+2018\right)^2}{x^2}\ge2017\Rightarrow A\ge\dfrac{2017}{2018}\)
Dấu "=" xảy ra <=> x = -2018
