Giải:
a) \(A=-x^2+6x+7\)
\(A=-x^2+6x-9+16\)
\(A=-\left(x^2-6x+9\right)+16\)
\(A=-\left(x-3\right)^2+16\le16;\forall x\)
\(\Rightarrow A_{Max}=16\)
\("="\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy ...
b) \(B=-2x^2+4x+5\)
\(\Leftrightarrow B=-2x^2+4x-2+7\)
\(\Leftrightarrow B=-\left(2x^2-4x+2\right)+7\)
\(\Leftrightarrow B=-\left(\sqrt{2}x-\sqrt{2}\right)^2+7\le7;\forall x\)
\(\Leftrightarrow B_{Max}=7\)
\("="\Leftrightarrow\sqrt{2}x-\sqrt{2}=0\Leftrightarrow x=1\)
Vậy ...
c) \(C=-3x^2+9x+8\)
\(\Leftrightarrow C=-3x^2+9x-\dfrac{27}{4}+\dfrac{59}{4}\)
\(\Leftrightarrow C=-\left(3x^2-9x+\dfrac{27}{4}\right)+\dfrac{59}{4}\)
\(\Leftrightarrow C=-\left(\sqrt{3}x-\dfrac{3\sqrt{3}}{2}\right)^2+\dfrac{59}{4}\le\dfrac{59}{4};\forall x\)
\(\Leftrightarrow C_{Max}=\dfrac{59}{4}\)
\("="\Leftrightarrow\sqrt{3}x-\dfrac{3\sqrt{3}}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy ...
