Đáp án:
$\min A = 2019\Leftrightarrow (x;y)=\left(-\dfrac12;0\right)$
Giải thích các bước giải:
$A = 2(x -y)^2 + 2(x +1)^2+ (y-1)^2+ 2017$
$\to A = 2(x^2 - 2xy + y^2) + 2(x^2 + 2x + 1) + (y^2 - 2y +1) + 2017$
$\to A = 4x^2 - 4xy + 3y^2 + 4x -2y + 2020$
$\to A = (4x^2- 4xy + y^2 + 4x - 2y + 1) + 2y^2 + 2019$
$\to A = (y - 2x -1)^2 + 2y^2 + 2019$
Ta có:
$\begin{cases}(y - 2x -1)^2 \geq 0\quad \forall x,y\\y^2 \geq 0\quad \forall y\end{cases}$
$\to (y - 2x -1)^2 + 2y^2 + 2019 \geq 2019$
$\to A \geq 2019$
Dấu $=$ xảy ra $\Leftrightarrow \begin{cases}y - 2x -1 = 0\\y = 0\end{cases}\Leftrightarrow \begin{cases}x = -\dfrac12\\y = 0\end{cases}$
Vậy $\min A = 2019\Leftrightarrow (x;y)=\left(-\dfrac12;0\right)$
