\(A=-x^2+x=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Dấu ''='' xảy ra khi x = 1/2
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\(A=-x^2+x=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Dấu '=' xảy ra khi x=1/2
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\(A=x\left(1-x\right)=x-x^2=-\left(x^2-x\right)=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
-Vậy \(A_{max}=\dfrac{1}{4}\)
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