\(A=x^2+y^2-4x+2y+7\)
\(=x^2+y^2-4x+2y+4+1+3\)
\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+3\)
\(=\left(x-2\right)^2+\left(y+1\right)^2+3\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+3\ge3>0\forall x,y\)
A= x2+y2-4x+2y+7
= (x2-4x+4)+(y2+2y+1)+2
= (x-2)2+(y+1)2+2
Ta thấy: (x-2)2\(\ge0\)
(y+1)
2\(\ge0\)
\(\Rightarrow\)(x-2)2+(y+1)2+2\(\ge2\)
\(\Rightarrow\)A\(\ge2\)
Vậy A>0 \(\forall x,y\)
\(A=x^2+y^2-4x+2y+7\)
\(=x^2+y^2-4x+2y+4+1+2\)
\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+2\)
\(=\left(x-2\right)^2+\left(y+1\right)^2+2\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+2\ge2>0\forall x,y\)
