Ta có: \(\sqrt{2x^2-4x+5}=\sqrt{2x^2-4x+2+3}=\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\)
Lại có: \(\left(\sqrt{2}x-\sqrt{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(\sqrt{2}x-\sqrt{2}\right)^2+3\ge3\)
\(\Rightarrow\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\ge\sqrt{3}\)
Vậy Min y là \(2+\sqrt{3}\)
\(y=2+\sqrt{2x^2-4x+5}=2+\sqrt{2x^2-4x+2+3}\)
\(=2+\sqrt{2\left(x^2-2x+1\right)+3}=2+\sqrt{2\left(x-1\right)^2+3}\)
Vì \(\left(x-1\right)^2\ge0\)\(\forall x\)
\(\Rightarrow2\left(x-1\right)^2\ge0\)\(\forall x\)\(\Rightarrow2\left(x-1\right)^2+3\ge3\)\(\forall x\)
\(\Rightarrow\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\)\(\forall x\)
\(\Rightarrow y=2+\sqrt{2\left(x-1\right)^2+3}\ge2+\sqrt{3}\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy \(miny=2+\sqrt{3}\)\(\Leftrightarrow x=1\)
