Tìm Min hoặc Max `a, A = x^2 + x` `b, B = ax^2 + bx + c` `c, C = 8x^2 + 12x + 50` 06/07/2021 Bởi Eloise Tìm Min hoặc Max `a, A = x^2 + x` `b, B = ax^2 + bx + c` `c, C = 8x^2 + 12x + 50`
a) $A = x^2 + x$ $\to A = \left(x + \dfrac{1}{2}\right)^2 – \dfrac{1}{4}$ $\to\min A = -\dfrac{1}{4}\Leftrightarrow x = -\dfrac{1}{2}$ b) $B = ax^2 + bx + c$ $\to B = a\left(x^2 + 2.\dfrac{b}{2a}x+ \dfrac{b^2}{4a^2}\right) + c -\dfrac{b^2}{4a}$ $\to B = a\left(x + \dfrac{b}{2a}\right)^2 + c -\dfrac{b^2}{4a}$ Với $a > 0$ $\to \min A = c -\dfrac{b^2}{4a}\Leftrightarrow x = -\dfrac{b}{2a}$ Với $a < 0$ $\to \max A = c -\dfrac{b^2}{4a}\Leftrightarrow x = -\dfrac{b}{2a}$ c) $C = 8x^2 + 12x + 50$ $\to C = 8\left(x +\dfrac{3}{4}\right)^2 +\dfrac{91}{2}$ $\to \min C = \dfrac{91}{2}\Leftrightarrow x = -\dfrac{3}{4}$ Bình luận
a) $A = x^2 + x$
$\to A = \left(x + \dfrac{1}{2}\right)^2 – \dfrac{1}{4}$
$\to\min A = -\dfrac{1}{4}\Leftrightarrow x = -\dfrac{1}{2}$
b) $B = ax^2 + bx + c$
$\to B = a\left(x^2 + 2.\dfrac{b}{2a}x+ \dfrac{b^2}{4a^2}\right) + c -\dfrac{b^2}{4a}$
$\to B = a\left(x + \dfrac{b}{2a}\right)^2 + c -\dfrac{b^2}{4a}$
Với $a > 0$
$\to \min A = c -\dfrac{b^2}{4a}\Leftrightarrow x = -\dfrac{b}{2a}$
Với $a < 0$
$\to \max A = c -\dfrac{b^2}{4a}\Leftrightarrow x = -\dfrac{b}{2a}$
c) $C = 8x^2 + 12x + 50$
$\to C = 8\left(x +\dfrac{3}{4}\right)^2 +\dfrac{91}{2}$
$\to \min C = \dfrac{91}{2}\Leftrightarrow x = -\dfrac{3}{4}$
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