Đáp án:
\(\min\left(2x^2 + 3y^2 - x + 2y - 1\right) = -\dfrac{35}{24}\Leftrightarrow (x;y)=\left(\dfrac14;-\dfrac13\right)\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad P = 2x^2 + 3y^2 - x + 2y - 1\\
\to P = 2\left(x^2 - \dfrac12x\right) + 3\left(y^2 + \dfrac23y\right) - 1\\
\to P = 2\left(x^2 - 2\cdot\dfrac14x + \dfrac{1}{16}\right) + 3\left(y^2 + 2\cdot\dfrac13y + \dfrac19\right)-\dfrac{35}{24}\\
\to P = 2\left(x - \dfrac14\right)^2 + 3\left(y + \dfrac13\right)^2 - \dfrac{35}{24}\\
\text{Ta có:}\\
\left(x - \dfrac14\right)^2\geqslant 0\quad \forall x\\
\left(y + \dfrac13\right)^2\geqslant 0\quad \forall y\\
\text{Do đó:}\\
2\left(x - \dfrac14\right)^2 + 3\left(y + \dfrac13\right)^2 - \dfrac{35}{24}\geqslant -\dfrac{35}{24}\quad \forall x,\ y\\
Hay\ P \geqslant -\dfrac{35}{24}\\
\text{Dấu = xảy ra}\ \Leftrightarrow \begin{cases}x - \dfrac14 = 0\\y + \dfrac13 = 0\end{cases}\Leftrightarrow \begin{cases}x = \dfrac14\\y = -\dfrac13\end{cases}\\
\text{Vậy}\ \min\left(2x^2 + 3y^2 - x + 2y - 1\right) = -\dfrac{35}{24}\Leftrightarrow (x;y)=\left(\dfrac14;-\dfrac13\right)
\end{array}\)
