\(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy Min A là : \(-3\Leftrightarrow x=2\)
\(B=4x^2+4x+11=\left(2x\right)^2+2.2x+1+10=\left(2x+1\right)^2+10\ge10\forall x\)Dấu " = " xảy ra
\(\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy Min B là : \(11\Leftrightarrow x=-\dfrac{1}{2}\)
\(A=x^2-4x+1\)
\(\Rightarrow A=x^2-4x+4-3\)
\(\Rightarrow A=\left(x-2\right)^2-3\)
Do \(\left(x-2\right)^2\ge0\) với \(\forall x\) (dấu "=" xảy ra \(\Leftrightarrow x-2=0\Rightarrow x=2\))
\(\Rightarrow\left(x-2\right)^2-3\ge-3\) hay \(A\ge-3\) (dấu "=" xảy ra \(\Leftrightarrow x=2\))
Vậy \(A_{min}=-3\) tại \(x=2\)
\(B=4x^2+4x+11\)
\(\Rightarrow B=\left(2x\right)^2+4x+1^2+10\)
\(\Rightarrow B=\left(2x+1\right)^2+10\)
Do \(\left(2x+1\right)^2\ge0\) với \(\forall x\) (dấu "=" xảy ra \(\Leftrightarrow2x+1=0\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\))
\(\Rightarrow\left(2x+1\right)^2+10\ge10\) hay \(B\ge10\) (dấu ''='' xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))
Vậy \(B_{min}=10\) tại \(x=\dfrac{1}{2}\)
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