a)Ta có : $x^2+3x+7=(x+\frac{3}{2})^2+\frac{19}{4}$$vì : (x+\frac{3}{2})^2\geq 0\Rightarrow (x+\frac{3}{2})^2+\frac{19}{4}\geq \frac{19}{4}\Rightarrow Min=\frac{19}{4}\Leftrightarrow x=\frac{-3}{2}$$b)-9x^2+12x-15=-(9x^2-12x+15)=-[(3x-2)^2+11]=-(3x-2)^2-11$$Vì :-(3x-2)^2\leq 0\Rightarrow -(3x-2)^2-11\leq -11\Rightarrow Max=-11\Leftrightarrow x=\frac{2}{3}$
a)Ta có : $x^2+3x+7=(x+\frac{3}{2})^2+\frac{19}{4}$$vì : (x+\frac{3}{2})^2\geq 0\Rightarrow (x+\frac{3}{2})^2+\frac{19}{4}\geq \frac{19}{4}\Rightarrow Min=\frac{19}{4}\Leftrightarrow x=\frac{-3}{2}$$b)-9x^2+12x-15=-(9x^2-12x+15)=-[(3x-2)^2+11]=-(3x-2)^2-11$$Vì :-(3x-2)^2\leq 0\Rightarrow -(3x-2)^2-11\geq -11\Rightarrow Max=-11\Leftrightarrow x=\frac{2}{3}$
a)Ta có : $x^2+3x+7=(x+\frac{3}{2})^2+\frac{19}{4}$$vì : (x+\frac{3}{2})^2\geq 0\Rightarrow (x+\frac{3}{2})^2+\frac{19}{4}\geq \frac{19}{4}\Rightarrow Min=\frac{19}{4}\Leftrightarrow x=\frac{-3}{2}$$b)-9x^2+12x-15=-(9x^2-12x+15)=-[(3x-2)^2+11]=-(3x-2)^2-11$$Vì :-(3x-2)^2\leq 0\Rightarrow -(3x-2)^2-11\
leq -11\Rightarrow Max=-11\Leftrightarrow x=\frac{2}{3}$