Đáp án:
$\min A = 2016 \Leftrightarrow (x;y)=(2;-2)$
Giải thích các bước giải:
Ta có:
$(x +\sqrt{x^2 + 2016})(y + \sqrt{y^2 + 2016})= 2016$
$\to \begin{cases}(x - \sqrt{x^2 + 2016})(x +\sqrt{x^2 + 2016})(y + \sqrt{y^2 + 2016})= 2016(x - \sqrt{x^2 + 2016})\\(x +\sqrt{x^2 + 2016})(y + \sqrt{y^2 + 2016})(y - \sqrt{y^2 + 2016})= 2016(y - \sqrt{y^2 + 2016})\end{cases}$
$\to \begin{cases}- 2016(y + \sqrt{y^2 + 2016}) = 2016(x - \sqrt{x^2 + 2016})\\-2016(x +\sqrt{x^2 + 2016}) = 2016(y - \sqrt{y^2 + 2016})\end{cases}$
$\to \begin{cases}y + \sqrt{y^2 + 2016} = \sqrt{x^2 + 2016} - x\\x +\sqrt{x^2 + 2016} = \sqrt{y^2 + 2016} - y\end{cases}$
$\to - x - y = x + y$
$\to y = - x$
Ta được:
$A = 2x^2 + xy - 4x + 2020$
$\to A = 2x^2 - x^2 - 4x + 2020$
$\to A = x^2 - 4x + 4 + 2016$
$\to A = (x - 2)^2 + 2016$
$\to A \geq 0 + 2016$
$\to A \geq 2016$
Dấu $=$ xảy ra $\Leftrightarrow x = 2 \to y = -2$
Vậy $\min A = 2016 \Leftrightarrow (x;y)=(2;-2)$
