Câu 1:

\(\text{ a) }E=x^2-2x+y^2+4y+8\\ E=x^2-2x+y^2+4y+1+4+3\\ E=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\\ E=\left(x-1\right)^2+\left(y+2\right)^2+3\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \left(y+2\right)^2\ge0\forall y\\ \Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x;y\\ \Rightarrow E=\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\forall x;y\\ \text{ Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\ \text{ Vậy }E_{\left(Min\right)}=3\text{ }khi\text{ }\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(\text{b) }F=x^2-4x+y^2-8y+6\\ F=x^2-4x+y^2-8y+4+16-14\\ F=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\\ F=\left(x-2\right)^2+\left(y-4\right)^2-14\\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \left(y-4\right)^2\ge0\forall y\\ \left(x-2\right)^2+\left(y-4\right)^2\ge0\forall x;y\\ F=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x;y\\ \text{Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\\ \text{Vậy }F_{\left(Min\right)}=-14\text{ }khi\text{ }\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)