\(A=2\left[x^2-\dfrac{1}{2}x+\dfrac{217}{2}\right]=2\left[x^2-\dfrac{1}{2}x+\dfrac{1}{16}+\dfrac{1735}{16}\right]\)
\(A=2\left[\left(x-\dfrac{1}{4}\right)^2+\dfrac{1735}{16}\right]\ge\dfrac{1735}{8}\) dấu"=" xảy ra<=>x=-1/4
\(A=2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)+\dfrac{1735}{8}=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{1735}{8}\ge\dfrac{1735}{8}\)
\(A_{min}=\dfrac{1735}{8}\) khi \(x=\dfrac{1}{4}\)
Ta có: \(A=2x^2-x+217\)
\(=2\left(x^2-\dfrac{1}{2}x+\dfrac{217}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1735}{16}\right)\)
\(=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{1735}{8}\ge\dfrac{1735}{8}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)
