\(a,A=5x-x^2\)
\(=-\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Vậy Max A = \(\dfrac{25}{4}\) khi \(x-\dfrac{5}{2}=0\Rightarrow x=\dfrac{5}{2}\)
\(b,B=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Vậy Max B = \(\dfrac{1}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(c,4x-x^2+3=7-\left(4-4x+x^2\right)\)
\(=7-\left(2-x\right)^2\le7\forall x\)
vậy Max C = 7 khi 2 - x =0 => x = 2
\(d,D=-x^2+8x-11=-\left(x^2-8x+16\right)+5\)
\(=-\left(x-4\right)^2+5\le5\forall x\)
vậy Max D = 5 khi x - 4 = 0 => x = 4
\(e,E=5-8x-x^2=21-\left(16+8x+x^2\right)\)
\(=21-\left(4+x\right)^2\le21\forall x\)
Vậy Max E = 21 khi 4 + x = 0 => x = -4
\(f,F=4x-x^2+1=5-\left(4-4x+x^2\right)\)
\(=5-\left(4-x\right)^2\le5\forall x\)
Vậy Max F = 5 khi 4 - x =0 => x = 4
