Đáp án:
\[{B_{\min }} = 2012 \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
B = \sqrt {{x^2} + {y^2} - 2xy + 2x - 2y + 5} + 2{y^2} - 8y + 2018\\
= \sqrt {\left( {{x^2} - 2xy + {y^2}} \right) + 2.\left( {x - y} \right) + 5} + \left( {2{y^2} - 8y + 8} \right) + 2010\\
= \sqrt {{{\left( {x - y} \right)}^2} + 2.\left( {x - y} \right) + 1 + 4} + 2.\left( {{y^2} - 4y + 4} \right) + 2010\\
= \sqrt {{{\left[ {\left( {x - y} \right) + 1} \right]}^2} + 4} + 2.{\left( {y - 2} \right)^2} + 2010\\
\ge \sqrt {0 + 4} + 2.0 + 2010 = 2012\\
\Rightarrow {B_{\min }} = 2012 \Leftrightarrow \left\{ \begin{array}{l}
{\left[ {\left( {x - y} \right) + 1} \right]^2} = 0\\
{\left( {y - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.
\end{array}\)
Vậy \({B_{\min }} = 2012 \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\)
