\(B=x^2-2xy+2y^2+2x-10y+17\\ \Rightarrow B=\left(x^2-2xy+y^2\right)+y^2+2x-10y+17\\ \Rightarrow B=\left[\left(x-y\right)^2+2\left(x-y\right)+1\right]+\left(y^2-8y+16\right)\\ \Rightarrow B=\left(x-y+1\right)^2+\left(y-4\right)^2\)
mà:\(\left\{{}\begin{matrix}\left(x-y+1\right)^2\ge0\\\left(y-4\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow B\ge0\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy \(B_{min}=0\Leftrightarrow\left(x;y\right)=\left(3;4\right)\)
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