Đáp án:

$MaxB = \dfrac{{36}}{5} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{7}{5};\dfrac{1}{5}} \right)$

Giải thích các bước giải:

$\begin{array}{l}
1)C = {x^2} + 5{y^2} + 3{z^2} - 4xy - 2yz - 2xz\\
 = {x^2} - 2x\left( {2y + z} \right) + {\left( {2y + z} \right)^2} - {\left( {2y + z} \right)^2} + 5{y^2} + 3{z^2} - 2yz\\
 = {\left( {x - 2y - z} \right)^2} + {y^2} - 6yz + 2{z^2}\\
 = {\left( {x - 2y - z} \right)^2} + {\left( {y - 3z} \right)^2} - 7{z^2}
\end{array}$

Bạn xem lại đề

$\begin{array}{l}
2)B =  - 2{x^2} - 3{y^2} + 3xy + 5x - 3y + 4\\
 =  - 2{x^2} + x\left( {3y + 5} \right) - 3{y^2} - 3y + 4\\
 =  - 2\left( {{x^2} - 2.x.\dfrac{{3y + 5}}{4} + {{\left( {\dfrac{{3y + 5}}{4}} \right)}^2}} \right) + 2{\left( {\dfrac{{3y + 5}}{4}} \right)^2} - 3{y^2} - 3y + 4\\
 =  - 2{\left( {x - \dfrac{{3y + 5}}{4}} \right)^2} - \dfrac{{15}}{8}{y^2} + \dfrac{3}{4}y + \dfrac{{57}}{8}\\
 =  - 2{\left( {x - \dfrac{{3y + 5}}{4}} \right)^2} - \dfrac{{15}}{8}\left( {{y^2} - 2.y.\dfrac{1}{5} + \dfrac{1}{{25}}} \right) + \dfrac{{36}}{5}\\
 =  - 2{\left( {x - \dfrac{{3y + 5}}{4}} \right)^2} - \dfrac{{15}}{8}{\left( {y - \dfrac{1}{5}} \right)^2} + \dfrac{{36}}{5}
\end{array}$

Lại có:

$\begin{array}{l}
{\left( {x - \dfrac{{3y + 5}}{4}} \right)^2} \ge 0,\forall x,y\\
{\left( {y - \dfrac{1}{5}} \right)^2} \ge 0,\forall y\\
 \Rightarrow  - 2{\left( {x - \dfrac{{3y + 5}}{4}} \right)^2} - \dfrac{{15}}{8}{\left( {y - \dfrac{1}{5}} \right)^2} \le 0,\forall x,y\\
 \Rightarrow  - 2{\left( {x - \dfrac{{3y + 5}}{4}} \right)^2} - \dfrac{{15}}{8}{\left( {y - \dfrac{1}{5}} \right)^2} + \dfrac{{36}}{5} \le \dfrac{{36}}{5},\forall x,y\\
 \Rightarrow B \le \dfrac{{36}}{5}\\
 \Rightarrow MaxB = \dfrac{{36}}{5}
\end{array}$

Dấu bằng xảy ra

$ \Leftrightarrow \left\{ \begin{array}{l}
x - \dfrac{{3y + 5}}{4} = 0\\
y - \dfrac{1}{5} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = \dfrac{1}{5}\\
x = \dfrac{7}{5}
\end{array} \right.$

Vậy $MaxB = \dfrac{{36}}{5} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{7}{5};\dfrac{1}{5}} \right)$