\(E=-2x^2+x=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)+\dfrac{1}{8}=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{1}{8}\le\dfrac{1}{8}\)\(E_{min}=\dfrac{1}{8}\) khi \(x=\dfrac{1}{4}\)Câu trả lời: \(E=-2x^2+x=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)+\dfrac{1}{8}=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{1}{8}\le\dfrac{1}{8}\)\(E_{min}=\dfrac{1}{8}\) khi \(x=\dfrac{1}{4}\)