a)Ta có : x2+3x+7=(x+32)2+194vì:(x+32)2≥0⇒(x+32)2+194≥194⇒Min=194⇔x=−32b)−9x2+12x−15=−(9x2−12x+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 :
x2+3x+7=(x+32)2+194vì:(x+32)2≥0⇒(x+32)2+194≥194⇒Min=194⇔x=−32b)−9x2+12x−15=−(9x2−12x+15)=−[(3x−2)2+11]=−(3x−2)2−11$Vì :-(3x-2)^2\leq 0\Rightarrow -(3x-2)^2-11\
leq -11\Rightarrow Max=-11\Leftrightarrow x=\frac{2}{3}$