Ta có:
\(A=2x^2+x+1\)
\(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)+\frac{7}{8}\)
\(=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\)
Vì \(\left(x+\frac{1}{4}\right)^2\ge0\) với mọi x nên \(2\left(x+\frac{1}{4}\right)^2\ge0\)
Suy ra \(A=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)
Vậy A đạt GTNN là \(\frac{7}{8}\) khi \(\left(x+\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow x+\frac{1}{4}=0\)
\(\Leftrightarrow x=-\frac{1}{4}\)
KL: Vậy A đạt GTNN là \(\frac{7}{8}\) khi \(x=-\frac{1}{4}\)