\(A=x^2y^2+24xy+144+2x^2+16x+32+15\)
\(=\left(xy+12\right)^2+2\left(x+4\right)^2+15\ge15\)
Dấu = xảy ra \(\Leftrightarrow x=-4;y=3\)
B=a2 +b2 +ab -3a -3b+2014
2B=2a2 +2b2 +2ab -6a -6b +4028
2B= (a+b)2 +(a-3)2 (b-3)2 + 4010
B=\(\frac{\left(a+b\right)^2}{2}+\frac{\left(a-3\right)^2}{2}+\frac{\left(b-3\right)^2}{2}+2005\) ≥2005
minB = 2005⇌\(\left[{}\begin{matrix}a=-b\\a=3\\b=3\end{matrix}\right.\)