Lời giải:
Ta có: \(Q=\frac{3}{2}x^2+x+1=(\sqrt{\frac{3}{2}}x)^2+2.\sqrt{\frac{3}{2}}x.\frac{\sqrt{6}}{6}+(\frac{\sqrt{6}}{6})^2+\frac{5}{6}\)
\(=\left(\sqrt{\frac{3}{2}}x+\frac{\sqrt{6}}{6}\right)^2+\frac{5}{6}\)
Vì \(\left(\sqrt{\frac{3}{2}}x+\frac{\sqrt{6}}{6}\right)^2\ge 0, \forall x\in\mathbb{R}\) nên \(Q\geq 0+\frac{5}{6}=\frac{5}{6}\)
Hay GTNN của \(Q=\frac{5}{6}\)
$Q$ đạt GTNN tại \(\left(\sqrt{\frac{3}{2}}x+\frac{\sqrt{6}}{6}\right)^2=0\Leftrightarrow x=-\frac{1}{3}\)
