\(A=x^2-4x+4-1=\left(x-2\right)^2-1\)
Do \(\left(x-2\right)^2\ge0;\forall x\)
\(\Rightarrow A\ge-1;\forall x\)
\(A_{min}=-1\) khi \(x-2=0\Rightarrow x=2\)
\(B=3\left(x^2+2.\dfrac{1}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x+\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(B_{min}=-\dfrac{4}{3}\) khi \(x=-\dfrac{1}{3}\)
\(C=-\left(x^2-6x+9\right)+24=-\left(x-3\right)^2+24\)
Do \(-\left(x-3\right)^2\le0;\forall x\Rightarrow C\le24\)
\(C_{max}=24\) khi \(x=3\)
\(D=x^2-2x+1+9-6x+x^2=2x^2-8x+10=2\left(x-2\right)^2+2\ge2\)
\(D_{min}=2\) khi \(x=2\)
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