\(A=-x^2+2x+3=-\left(x^2-2x-3\right)\)
\(=-\left(x^2-2x+1-4\right)\)
\(=-\left[\left(x-1\right)^2-4\right]=-\left(x-1\right)^2+4\le4\)
Vậy \(A_{max}=4\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(B=-2x^2-4x=-2\left(x^2+2x\right)\)
\(=-2\left(x^2+2x+1-1\right)\)
\(=-2\left[\left(x+1\right)^2-1\right]=-\left(x+1\right)^2+2\le2\)
Vậy \(B_{max}=2\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
\(C=-x^2-6x+12=-\left(x^2+6x-12\right)\)
\(=-\left(x^2+6x+9-21\right)\)
\(=-\left[\left(x+3\right)^2-21\right]=-\left(x+3\right)^2+21\le21\)
Vậy \(C_{max}=21\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(D=-x^2+3x-1==-\left(x^2-3x+1\right)\)
\(=-\left(x^2-3x+\frac{9}{4}-\frac{5}{4}\right)\)
\(=-\left[\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right]=-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)
Vậy \(D_{max}=\frac{5}{4}\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
\(E=-x^2-5x=-\left(x^2+5x\right)\)
\(=-\left(x^2+5x+\frac{25}{4}-\frac{25}{4}\right)\)
\(=-\left[\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\right]\)
\(=-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Vậy \(E_{max}=\frac{25}{4}\Leftrightarrow x+\frac{5}{2}\Leftrightarrow x=-\frac{5}{2}\)
\(F=-3x^2+12x-3=-3\left(x^2-4x+1\right)\)
\(=-3\left(x^2-4x+4-3\right)\)
\(=-3\left[\left(x-2\right)^2-3\right]\le9\)
Vậy \(F_{min}=9\Leftrightarrow x-2=0\Leftrightarrow x=2\)
