\(B=x^2-2x+y^2-4y+6.\)
\(=x^2-2x+1+y^2-4y+4+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1\)
Vì \(\left(x+1\right)^2\ge0\)và \(\left(y-2\right)^2\ge0\)
\(\Rightarrow B_{min}\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)
\(\Rightarrow B_{min}=1\)\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)