\(A=x^2-x+2009\)
\(=x^2-x+\frac{1}{4}+2008,75\)
\(=\left(x-\frac{1}{2}\right)^2+2008,75\)
\(\left(x-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+2008,75\ge2008,75\)
Dấu ''='' xảy ra khi \(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(MinA=2008,75\Leftrightarrow x=\frac{1}{2}\)
Ta có :
\(x^2-x+2009\)
\(=x^2-2.x.\frac{1}{2}+\frac{1}{4}+2009-\frac{1}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{8035}{4}\)
\(\left(x-\frac{1}{2}\right)^2+\frac{8035}{4}\ge\frac{8035}{4}\forall x\)
Dấu " = " xảy ra khi x = 1/2
Vậy ......
