N = x2 + x + 1
= x2 + 2.x.\(\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
= \(\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\)với mọi x
\(\Rightarrow\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\)
hay \(N\ge\frac{-1}{4}\)
Dấu " = " xảy ra <=> \(x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)
Vậy GTNN của \(N=\frac{-1}{4}\Leftrightarrow x=\frac{-1}{2}\)
Bài của NGUYỄN VĂN HUY sai nhé
\(N=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra <=> \(x=-\frac{1}{2}\)
Vậy MIN \(N=\frac{3}{4}\) khi \(x=-\frac{1}{2}\)
