a) A=x2+8x
=x2+8x+16-16
=(x+4)2-16
Vì \(\left(x+4\right)^2\ge0\forall x\in R\Rightarrow A\ge-16\)
MinA=-16 khi (x+4)2=0 <=> x+4=0 <=> x=-4
b) B=-2x2+8x-15
= -(2x2-8x+15)
=-\(\left[\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.2\sqrt{2}+\left(2\sqrt{2}\right)^2-\left(2\sqrt{2}\right)^2+15\right]\)
=\(-\left[\left(\sqrt{2}x-2\sqrt{2}\right)^2+7\right]\)
\(=-\left(\sqrt{2}x-2\sqrt{2}\right)^2-7\le7\)
MaxB=-7 khi x=2
b) \(-2x^2+8x-15=-2\left(x^2-4x+4\right)-7=-2\left(x-2\right)^2-7\le-7\)
Đẳng thức xảy rA \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)