a) \(A=-x^2+4x+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\ge7\)
Dấu "=" xảy ra khi và chỉ khi x = 2
Vậy Max A = 7 <=> x = 2
b) \(B=-x^2+x=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra khi và chỉ khi x = \(\frac{1}{2}\)
Vậy Max B = \(\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)
c) \(C=-2x^2+2x-5=-2\left(x^2-x\right)-5=-2\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{2}-5\)
\(=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\le-\frac{9}{2}\)
Dấu "=" xảy ra khi và chỉ khi x = \(\frac{1}{2}\)
Vậy Max C = \(-\frac{9}{2}\Leftrightarrow x=\frac{1}{2}\)
\(a,A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\) Vậy \(Max_A=7\) khi \(x-2=0\Rightarrow x=2\)
\(b,x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)Vậy \(Max_B=\dfrac{1}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(c,2x-2x^2+5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-\left(x-\dfrac{1}{2}\right)-\dfrac{9}{2}\le\dfrac{-9}{2}\)Vậy \(Max_C=\dfrac{-9}{2}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
