Giải:
1) \(A=x^2-6x+15\)
\(\Leftrightarrow A=x^2-6x+9+6\)
\(\Leftrightarrow A=\left(x^2-6x+9\right)+6\)
\(\Leftrightarrow A=\left(x-3\right)^2+6\ge6;\forall x\)
\(\Leftrightarrow A_{Min}=6\)
\("="\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy ...
2) \(B=-3x^2+2x-6\)
\(\Leftrightarrow B=-3x^2+2x-\dfrac{1}{3}-\dfrac{17}{3}\)
\(\Leftrightarrow B=-\left(3x^2-2x+\dfrac{1}{3}\right)-\dfrac{17}{3}\)
\(\Leftrightarrow B=-\left(\sqrt{3}x-\dfrac{1}{\sqrt{3}}\right)^2-\dfrac{17}{3}\)
\(\Leftrightarrow B=-\dfrac{17}{3}-\left(\sqrt{3}x-\dfrac{1}{\sqrt{3}}\right)^2\le-\dfrac{17}{3};\forall x\)
\(\Leftrightarrow B_{Max}=-\dfrac{17}{3}\)
\("="\Leftrightarrow\sqrt{3}x-\dfrac{1}{\sqrt{3}}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy ...
