\(N=x^2-x=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\) hay \(N\ge-\dfrac{1}{4}\)
Dấu ''='' xảy ra \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
Vậy \(minN=-\dfrac{1}{4}\) khi \(x=\dfrac{1}{2}\)
