\(B=x-x^2\)
\(=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left[x^2-2.x\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Ta có :
\(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Dấu = xảy ra \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(Max_B=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
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