Đáp án:
$minA = \dfrac{8}{3} \Leftrightarrow (x;y) = \left\{\left(\dfrac{2\sqrt3}{3};-\dfrac{2\sqrt3}{3}\right),\left(-\dfrac{2\sqrt3}{3};\dfrac{2\sqrt3}{3}\right)\right\}$
$maxA = 8 \Leftrightarrow x = y = \pm 2$
Giải thích các bước giải:
Ta có:
$+) \, (x - y)^2 \geq 0$
$\Leftrightarrow x^2 + y^2 \geq 2xy$
$\Leftrightarrow \dfrac{x^2 + y^2}{2} \geq xy$
$+) \, (x + y)^2 \geq 0$
$\Leftrightarrow x^2 + y^2 \geq - 2xy$
$\Leftrightarrow \dfrac{x^2 + y^2}{2} \geq -xy$
$\Leftrightarrow - \dfrac{x^2 + y^2}{2} \leq xy$
Ta được:
$+) \, A = x^2 + y^2 = 4 + xy$
$\Leftrightarrow x^2 + y^2 \leq 4 + \dfrac{x^2 + y^2}{2}$
$\Leftrightarrow 2(x^2 + y^2) \leq 8 + x^2 + y^2$
$\Leftrightarrow x^2 + y^2 \leq 8$
Hay $maxA = 8$
Dấu = xảy ra $\Leftrightarrow x = y = \pm 2$
$+) \, A = x^2 + y^2 = 4 + xy$
$\Leftrightarrow x^2 + y^2 \geq 4 - \dfrac{x^2 + y^2}{2}$
$\Leftrightarrow 2(x^2 + y^2) \geq 8 - (x^2 + y^2)$
$\Leftrightarrow 3(x^2 + y^2) \geq 8$
$\Leftrightarrow x^2 + y^2 \geq \dfrac{8}{3}$
Hay $minA = \dfrac{8}{3}$
Dấu = xảy ra $\Leftrightarrow x = - y \Leftrightarrow (x;y) = \left\{\left(\dfrac{2\sqrt3}{3};-\dfrac{2\sqrt3}{3}\right),\left(-\dfrac{2\sqrt3}{3};\dfrac{2\sqrt3}{3}\right)\right\}$
