\(HIEU3D=-x^2+6x-10\)
\(HIEU3D=-1\left(x^2-6x+10\right)\)
\(HIEU3D=-1\left(x^2-6x+9+1\right)\)
\(HIEU3D=-1\left(x^2-6x+9\right)-1\)
\(HIEU3D=-1\left(x-3\right)^2-1\le-1\)
Dấu "=" xảy ra khi: \(x=3\)
\(-x^2+6x-10=-\left(x^2-6x+10\right)\)
\(=-\left(x^2-6x+9+1\right)=-\left(x^2-6x+9\right)-1\)\(=-\left(x-3\right)^2-1\)
Do \(-\left(x-3\right)^2\le0\left(\forall x\right)\)
\(\Rightarrow-\left(x-3\right)^2-1\le-1\left(\forall x\right)\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow x=3\)
A=x2+6x-10 = x2+6x+9-19 = (x-3)2 - 15
ta có (x-3)2 \(\ge\)0
=> (x-3)2 - 15 \(\ge\) -15
A= -15 khi x= 3
B=\(-x^2+6x-10\)
=\(-\left(x^2-2.x.3+3^2\right)+1\)
=\(-\left(x+3\right)^2+1\)
Vậy GTLM của bt là 1
