Đáp án:
\[\left\{ \begin{array}{l}
{\left( {x + y} \right)_{\max }} = \sqrt 2 \Leftrightarrow x = y = \frac{{\sqrt 2 }}{2}\\
{\left( {x + y} \right)_{\min }} = - \sqrt 2 \Leftrightarrow x = y = \frac{{ - \sqrt 2 }}{2}
\end{array} \right.\]
Giải thích các bước giải:
\(\begin{array}{l}
{\left( {x - y} \right)^2} \ge 0\\
\Leftrightarrow {x^2} - 2xy + {y^2} \ge 0\\
\Leftrightarrow {x^2} + {y^2} \ge 2xy\\
\Leftrightarrow 2\left( {{x^2} + {y^2}} \right) \ge {x^2} + 2xy + {y^2}\\
\Leftrightarrow 2 \ge {\left( {x + y} \right)^2}\\
\Leftrightarrow {\left( {x + y} \right)^2} \le 2\\
\Leftrightarrow - \sqrt 2 \le x + y \le \sqrt 2 \\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + y} \right)_{\max }} = \sqrt 2 \Leftrightarrow x = y = \frac{{\sqrt 2 }}{2}\\
{\left( {x + y} \right)_{\min }} = - \sqrt 2 \Leftrightarrow x = y = \frac{{ - \sqrt 2 }}{2}
\end{array} \right.
\end{array}\)
