a)x2+ 5x+8
\(=\left(x^2+5x+\frac{25}{4}+\frac{7}{4}\right)\)
\(=\left(x^2+5x+\left(\frac{5}{2}\right)^2\right)+\frac{7}{4}\)
\(=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}\ge0+\frac{7}{4}=\frac{7}{4}\)
Dấu = khi \(x=-\frac{5}{2}\)
Vậy \(Min_A=\frac{7}{4}\Leftrightarrow x=-\frac{5}{2}\)
\(x^2+5x+8\)
\(=x^2+2x\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+8\)
\(=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}\)
Vì \(\left(x+\frac{5}{2}\right)^2\ge0\)
Nên \(\left(x+\frac{5}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
Vậy GTNN của đa thức là: \(\frac{7}{4}\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)
