\(A=-x^2-4x-4-9y^2-6y-1+8\)
\(=-\left(x+2\right)^2-\left(3y+1\right)^2+8< =8\)
Dấu '=' xảy ra khi x=-2 và y=-1/3
\(B=-2\left(x^2-\dfrac{1}{2}x+1\right)\)
\(=-2\left(x^2-2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{15}{16}\right)\)
\(=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{15}{8}< =-\dfrac{15}{8}\)
Dấu = xảy ra khi x=1/4