A=4x-x2
\(=4-4+4x^2-x^2\)
\(=4-\left(x^2-4x+4\right)\)
\(=4-\left(x-2\right)^2\le4\)
Dấu = khi x=2
Vậy MaxA=4 khi x=2
b)-9x2+6x-2
\(=-9\left(x^2+\frac{2x}{3}+\frac{1}{9}\right)-1\)
\(=-9\left(x-\frac{1}{3}\right)^2-1\le-1\)
Dấu = khi \(x=\frac{1}{3}\)
Vậy MaxB=-1 khi \(x=\frac{1}{3}\)
\(A=4x-x^2=4-x^2+4x-4=4-\left(x-2\right)^2\ge4\)
\(M\text{ax}A=4\Leftrightarrow x=2\)
\(B=-9x^2+6x-2=-1-9x^2+6x-1=-1-\left(3x-1\right)^2\ge-1\)
\(M\text{ax}B=-1\Leftrightarrow x=\frac{1}{3}\)