Ta có:
\(x^2+x\sqrt{3}+1\)
\(=x^2+2x.\dfrac{\sqrt{3}}{2}+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x+\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Ta lại có: \(\left(x+\dfrac{\sqrt{3}}{2}\right)^2\ge0\)
\(\Rightarrow\left(x+\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\dfrac{\sqrt{3}}{2}\right)^2=0\)
\(\Leftrightarrow x+\dfrac{\sqrt{3}}{2}=0\)
\(\Leftrightarrow x=-\dfrac{\sqrt{3}}{2}\)
Vậy Min\(x^2+x\sqrt{3}+1=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{\sqrt{3}}{2}\)
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