[toan 9] tìm cực trị

C

chaugiang81


ĐKXĐ: a khác 1
$ \dfrac{1}{2(1+\sqrt{a})} + \dfrac{1}{2(1-\sqrt{a})} - \dfrac{a^2 +2}{1-a^3}$
$= \dfrac{1 - \sqrt{a}}{2(1-a)} + \dfrac{1 + \sqrt{a}}{2(1-a)} - \dfrac{a^2 + 2}{(1-a)(a^2 +a +1)}$
$= \dfrac{1}{1-a}- \dfrac{a^2 +2}{(1-a)(a^2 +a +1)}$
$= \dfrac{a^2 + a+1-a^2 -2}{-(a-1)(a^2 +a+1)}$
$= \dfrac{a-1}{-(a-1)(a^2 +a+1)}$
$= -\dfrac{1}{a^2 +a +1}$

 
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E

eye_smile

b,

$a^2+a+1 \ge 1$ do $a \ge 0;a$ khác 1

\Rightarrow $\dfrac{1}{a^2+a+1} \le 1$

\Rightarrow $-\dfrac{1}{a^2+a+1} \ge -1$

Dấu = xảy ra \Leftrightarrow $a=0$
 
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