Lời giải:
Ta có:
\(A=3-(10x^2+4xy+4y^2)=3-[9x^2+(x^2+4xy+4y^2)]\)
\(=3-[(3x)^2+(x+2y)^2]\)
Vì \((3x)^2\geq 0; (x+2y)^2\geq 0\Rightarrow (3x)^2+(x+2y)^2\geq 0, \forall x,y\)
\(\Rightarrow A=3-[(3x)^2+(x+2y)^2]\leq 3\)
Vậy $A_{\max}=3$. Dấu "=" xảy ra khi \((3x)^2=(2x+y)^2=0\Rightarrow x=y=0\)