Lời giải:
a) \(A=x^2+2y^2-2xy+2x-10y\)
\(\Leftrightarrow A=(x-y+1)^2+(y-4)^2-17\)
Ta thấy \((x-y+1)^2; (y-4)^2\geq 0\Rightarrow A\geq -17\)
Vậy \(A_{\min}=-17\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x-y+1=0\\ y-4=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=3\\ y=4\end{matrix}\right.\)
b)
\(B=x^2+6y^2+14z^2-8yz+6xz-4xy\)
\(\Leftrightarrow B=(x-2y+3z)^2+2y^2+5z^2+4yz\)
\(\Leftrightarrow B=(x-2y+3z)^2+2(y+z)^2+z^2\)
Ta thấy \((x-2y+3z)^2; (y+z)^2; z^2\geq 0\forall x,y,z\in\mathbb{R}\)
\(\Rightarrow B\geq 0\Leftrightarrow B_{\min}=0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x-2y+3z=0\\ y+z=0\\ z=0\end{matrix}\right.\Leftrightarrow x=y=z=0\)
