toán 9- bài toán cực trị

V

vipboycodon

Theo bđt bunhia ta có:
$P = \sqrt{x-5}+\sqrt{13-x} \le \sqrt{2(x-5+13-x)} = 4$
Dấu "=" xảy ra khi $x = 9$
 
H

huynhbachkhoa23

Đặt $\sqrt{x-5}=u$ và $\sqrt{13-x}=v$

$u^2+v^2=8$

Đặt $u=2\sqrt{2}\sin t; v= 2\sqrt{2}\cos t$ với $t\in \left [0; \dfrac{\pi}{2}\right ]$

$\sqrt{x-5}+\sqrt{13-x}=2\sqrt{2}(\sin t + \cos t)=4\sin \left (t+\dfrac{\pi}{4} \right ) \le 4$

$\text{max P}=4 \leftrightarrow \sin \left (t+\dfrac{\pi}{4} \right )=1$

$\leftrightarrow t=\dfrac{\pi}{4} \leftrightarrow 2\sqrt{2}\sin t = 2 \to u=2\to x=9$
 
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