\(D=\left(x-1\right)^2+\left(x-3\right)^2\)
\(=x^2-2x+1+x^2-6x+9\)
\(=2x^2-8x+10\)
\(=2\left(x^2-4x+5\right)\)
\(=2\left[\left(x^2-4x+4\right)+1\right]\)
\(=2\left[\left(x-2\right)^2+1\right]\)
\(=2\left(x-2\right)^2+2\)
Ta có : \(\left(x-2\right)^2\ge0\forall x\Rightarrow2\left(x-2\right)^2\ge0\Rightarrow2\left(x-2\right)^2+2\ge2\)
hay D ≥ 2
Dấu = xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(Min_D=2\Leftrightarrow x=2\)
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