Gtnn

I

iceghost

3) ĐKXĐ : $5 \le x \le 23$

Đặt $P=\sqrt{x-5} + \sqrt{23-x}$
$\implies P^2 = (\sqrt{x-5} + \sqrt{23-x})^2$
Áp dụng bđt Bunhiacopxki ta có :
$P^2 \le (1^2+1^2)[(\sqrt{x-5})^2 + (\sqrt{23-x})^2] = 2.18 = 36 \\
\implies P \le 6$
Vậy $Max_P = 6 \iff \dfrac{\sqrt{x-5}}1=\dfrac{\sqrt{23-x}}1 \iff x-5 = 23-x \iff x=14$
 
You must log in or register to reply here.
Top Bottom