Đáp án:
$\min(x^2 + y^2 - xy - x + y +1)=\dfrac23\Leftrightarrow (x;y)=\left(\dfrac13;-\dfrac13\right)$
Giải thích các bước giải:
$x^2 + y^2 - xy - x + y +1$
$= x^2 + \dfrac14y^2 + \dfrac14 - 2\cdot x\cdot\dfrac12y - 2\cdot x\cdot\dfrac12 + 2\cdot \dfrac12y\cdot\dfrac12 +\dfrac34y^2 +\dfrac12y + \dfrac34$
$= \left(x - \dfrac12y -\dfrac12\right)^2 + \dfrac34\left(y^2 + 2\cdot\dfrac13\cdot y + \dfrac19\right) +\dfrac23$
$= \left(x - \dfrac12y -\dfrac12\right)^2 + \dfrac34\left(y +\dfrac13\right)^2 +\dfrac23$
Ta có:
$\quad \begin{cases}\left(x - \dfrac12y -\dfrac12\right)^2 \geq 0\quad \forall x;\, y\\\dfrac34\left(y +\dfrac13\right)^2 \geq 0\quad \forall y\end{cases}$
$\to \left(x - \dfrac12y -\dfrac12\right)^2 + \dfrac34\left(y +\dfrac13\right)^2 +\dfrac23 \geq \dfrac23$
Dấu $=$ xảy ra $\Leftrightarrow \begin{cases}x - \dfrac12y -\dfrac12 = 0\\y +\dfrac13 = 0\end{cases}\Leftrightarrow \begin{cases}x =\dfrac13\\y=-\dfrac13\end{cases}$
Vậy $\min(x^2 + y^2 - xy - x + y +1)=\dfrac23\Leftrightarrow (x;y)=\left(\dfrac13;-\dfrac13\right)$
