Đáp án:
$\min H = -\dfrac12 \Leftrightarrow x = 2$
$\max H = 2 \Leftrightarrow x = -\dfrac12$
Giải thích các bước giải:
$\quad H =\dfrac{3 - 4x}{2x^2 + 2}$
$+)\quad \min$
$\quad H +\dfrac12 =\dfrac{3 - 4x}{2x^2 + 2} + \dfrac12$
$\to H +\dfrac12 =\dfrac{2(3 - 4x) + 2x^2 + 2}{2(2x^2 +2)}$
$\to H +\dfrac12 =\dfrac{2x^2 - 8x + 8}{2(2x^2 +2)}$
$\to H +\dfrac12 =\dfrac{(x-2)^2}{2x^2 + 2}$
$\to H +\dfrac12\geqslant 0$
$\to H \geqslant -\dfrac12$
Dấu $=$ xảy ra $\Leftrightarrow x - 2 = 0\Leftrightarrow x = 2$
$+)\quad \max$
$\quad H - 2 = \dfrac{3 - 4x}{2x^2 + 2}- 2$
$\to H - 2 =\dfrac{3 - 4x - 2(2x^2 +2)}{2x^2 + 2}$
$\to H - 2 =\dfrac{- 4x^2 - 4x - 1}{2x^2 +2}$
$\to H-2 = -\dfrac{(2x+1)^2}{2x^2 +2}$
$\to H - 2 \leqslant 0$
$\to H \leqslant 2$
Dấu $=$ xảy ra $\Leftrightarrow 2x + 1 = 0 \Leftrightarrow x = -\dfrac12$
Vậy $\min H = -\dfrac12 \Leftrightarrow x = 2$
$\max H = 2 \Leftrightarrow x = -\dfrac12$
