\(A=2x^2-3y+8x+y^2+11\)
\(=\left(2x^2+8x+8\right)+\left(y^2-3y+\frac{9}{4}\right)+\frac{3}{4}\)
\(=2\left(x^2+4x+4\right)+\left(y^2-3y+\frac{9}{4}\right)+\frac{3}{4}\)
\(=2\left(x+2\right)^2+\left(y-\frac{3}{2}\right)^2+\frac{3}{4}\)
Vì: \(2\left(x+2\right)^2+\left(y-3\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x,y\)
\(\Rightarrow2\left(x+2\right)^2+\left(y-\frac{3}{2}\right)^2+\frac{3}{4}>0\forall x,y\)
=.= hok tốt!!
Ta có\(A=2x^2-3y+8x+y^2+11\)
\(=2.\left(x^2+2.x.4+4^2\right)-5-3y+y^2\)
\(=2.\left(x+4\right)^2+\left(y^2-2.y.\frac{3}{2}+\frac{9}{4}\right)-5-\frac{9}{4}\)
\(=2.\left(x+4\right)^2+\left(y-\frac{3}{2}\right)^2-\left(5+\frac{9}{4}\right)< 0\)với mọi x
Không thể làm luôn dương được , chắc mình sai , thôi góp ý vậy
