Đáp án:
a. Max=2017
Giải thích các bước giải:
\(\begin{array}{l}
a. A = - 2{x^2} - 10{y^2} + 4xy + 4x + 4y + 2013\\
= - 2\left( {{x^2} + {y^2} + 1 - 2xy - 2x + 2y} \right) - 8{y^2} - 8y + 2015\\
= - 2{\left( { - x + y + 1} \right)^2} - 8\left( {{y^2} + 2.y.\dfrac{1}{2} + \dfrac{1}{4}} \right) + 2017\\
= - 2{\left( { - x + y + 1} \right)^2} - 8{\left( {y + \dfrac{1}{2}} \right)^2} + 2017\\
Do:\left\{ \begin{array}{l}
{\left( { - x + y + 1} \right)^2} \ge 0\\
{\left( {y + \dfrac{1}{2}} \right)^2} \ge 0
\end{array} \right.\forall x;y\\
\to \left\{ \begin{array}{l}
- 2\left( { - x + y + 1} \right) \le 0\\
- 8{\left( {y + \dfrac{1}{2}} \right)^2} \le 0
\end{array} \right.\\
\to - 2{\left( { - x + y + 1} \right)^2} - 8{\left( {y + \dfrac{1}{2}} \right)^2} \le 0\\
\to - 2{\left( { - x + y + 1} \right)^2} - 8{\left( {y + \dfrac{1}{2}} \right)^2} + 2017 \le 2017\\
\to Max = 2017\\
\Leftrightarrow \left\{ \begin{array}{l}
- x + y + 1 = 0\\
y + \dfrac{1}{2} = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - \dfrac{1}{2}\\
x = \dfrac{1}{2}
\end{array} \right.\\
b.B = \dfrac{3}{{2{x^2} + 2x + 3}} = \dfrac{3}{{2{x^2} + 2.x\sqrt 2 .\dfrac{1}{2} + \dfrac{5}{2}}}\\
= \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{2}} \right)}^2} + \dfrac{5}{2}}}\\
Do:{\left( {x\sqrt 2 + \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 2 + \dfrac{1}{2}} \right)^2} + \dfrac{5}{2} \ge \dfrac{5}{2}\\
\to \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{2}} \right)}^2} + \dfrac{5}{2}}} \le 3:\dfrac{5}{2}\\
\to \dfrac{3}{{{{\left( {x\sqrt 2 + \dfrac{1}{2}} \right)}^2} + \dfrac{5}{2}}} \le \dfrac{6}{5}\\
\to Max = \dfrac{6}{5}\\
\Leftrightarrow x\sqrt 2 + \dfrac{1}{2} = 0\\
\to x = - \dfrac{1}{2}
\end{array}\)
