\(x^2+x+2\)
\(=x^2+x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(x+\dfrac{1}{2}=0\)
=>\(x=-\dfrac{1}{2}\)
\(x^2+x+2\)
\(=\left[x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\dfrac{1}{4}+2\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)
Ta thấy: \(\left(x+\dfrac{1}{2}\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4};\forall x\)
Dấu \("="\) xảy ra khi: \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy GTNN của biểu thức đã cho bằng \(\dfrac{7}{4}\) tại \(x=-\dfrac{1}{2}\).
