a, \(A=x^2-2x+3=\left(x^2-2x+1\right)+2\)
\(=\left(x-1\right)^2+2\ge2\forall x\)
Vậy MinA = 2 khi \(x-1=0\Rightarrow x=1\)
\(b,x^2+8x+20=\left(x^2+8x+16\right)+4\)
\(=\left(x+4\right)^2+4\ge4\forall x\)
Vậy Min B = 4 khi \(x+4=0\Rightarrow x=-4\)
\(c,C=x^2+x+1=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Vậy Min C = \(\dfrac{3}{4}\) khi \(x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)
\(d,D=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{4}\)\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
Vậy Min D = \(\dfrac{11}{4}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
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