đặt A=\(x^2+x\sqrt{3}+1\)
= \(x^2+2x.\dfrac{\sqrt{3}}{2}+\dfrac{3}{4}+\dfrac{1}{4}\)
= \(\left(x^2+2x.\dfrac{\sqrt{3}}{2}+\dfrac{3}{4}\right)+\dfrac{1}{4}\)
= \(\left(x+\dfrac{3}{4}\right)^2+\dfrac{1}{4}\)
do \(\left(x+\dfrac{3}{4}\right)^2\ge0\) ∀ x
⇔ \(\left(x+\dfrac{3}{4}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
⇔ A \(\ge\dfrac{1}{4}\)
=> Min A = \(\dfrac{1}{4}\) dấu "=" xảy ra khi x= \(\dfrac{-3}{4}\)
Giải:
Đặt \(A=x^2+x\sqrt{3}+1\)
\(\Leftrightarrow A=x^2+2.x\dfrac{\sqrt{3}}{2}+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(\Leftrightarrow A=\left(x+\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Vì \(\left(x+\dfrac{\sqrt{3}}{2}\right)^2\ge0;\forall x\)
\(\Leftrightarrow\left(x+\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4};\forall x\)
\(\Leftrightarrow A\ge\dfrac{1}{4};\forall x\)
\(\Leftrightarrow A_{Min}=\dfrac{1}{4}\)
\(\Leftrightarrow x+\dfrac{\sqrt{3}}{2}=0\Leftrightarrow x=-\dfrac{\sqrt{3}}{2}\)
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