a: \(A=-4x^2+4x-1\)
\(=-\left(4x^2-4x+1\right)\)
\(=-\left(2x-1\right)^2\le0\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
b: \(B=-x^2+5x\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)
a) \(A=-4x^2+4x-1=-\left(4x^2-4x+1\right)\)
\(=-\left(2x-1\right)^2\le0\)
\(maxA=0\Leftrightarrow x=\dfrac{1}{2}\)
b) \(B=-x^2+5x=-\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
\(maxB=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)
c) \(C=-3x^2-9x+6=-3\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{51}{4}\)
\(=-3\left(x+\dfrac{3}{2}\right)^2+\dfrac{51}{4}\le\dfrac{51}{4}\)
\(maxC=\dfrac{51}{4}\Leftrightarrow x=-\dfrac{3}{2}\)