Ta có: \(x-x^2=-x^2+x\)
\(=-x^2+x-\dfrac{1}{4}+\dfrac{1}{4}\)
\(=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Vì: \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu = xảy ra khi : \(-\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
Vậy max =1/4 tại x=1/2
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