\(A=x^2+x+2018\)
\(A=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+2018\)
\(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{8071}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{8071}{4}\ge\dfrac{8071}{4}\)
\(\Rightarrow\) \(Amin=\dfrac{8071}{4}\Leftrightarrow x+\dfrac{1}{2}=0\)
\(\Rightarrow x=-\dfrac{1}{2}\)
Vậy \(Amin=\dfrac{8071}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
Ta có : \(x^2+x+2018\)
\(=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{8071}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{8071}{4}\ge\dfrac{8071}{4}\)
Vật MIN của biểu thức là \(\dfrac{8071}{4}\) khi \(\left(x+\dfrac{1}{2}\right)^2\ge0\Leftrightarrow x=-\dfrac{1}{2}\)
