Question

Tìm giá trị nhỏ nhất

x² + x + 1

Answer 1

\(A=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow A_{Min}=\dfrac{3}{4}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{-1}{2}\)

Vậy, A đạt GTNN \(\Leftrightarrow x=\dfrac{-1}{2}\)