\(Ta có:A= x^2 +x+\dfrac{1}{2}\)
\(=x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\bigg(x+2\bigg)^2+\dfrac{1}{4}\)
\(Vì: \bigg(x+\dfrac{1}{2}\bigg)^2\ge0 \)
\(\Rightarrow\bigg(x+\dfrac{1}{2}\bigg)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
Vậy giá trị nhỏ nhất của A=x2+x+1/2 là 1/4 khi \(\bigg(x+\dfrac{1}{2}\bigg)^2=0\)
\(\Leftrightarrow x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)