tìm GÍA TRỊ LỚN NHẤT

D

demon311

$\sqrt[]{ 3-x}+x = 2.\dfrac{ 1}{2}\sqrt[]{ 3-x}+x \le \dfrac{ 1}{4}+3-x+x=\dfrac{ 13}{4}$

Dấu bằng: $\sqrt[]{ 3-x}=\dfrac{ 1}{2}$
 
Last edited by a moderator:
V

vipboycodon


Cách khác: $A = x+\sqrt{3-x}$
Đặt $\sqrt{3-x} = t \ge 0$
$\rightarrow 3-x = t^2 \rightarrow x = 3-t^2$
$\rightarrow A = 3-t^2+t = -(t-\dfrac{1}{2})^2+\dfrac{13}{4} \le \dfrac{13}{4}$
Vậy Max $A = \dfrac{13}{4}$ khi $t = \dfrac{1}{2} \rightarrow x = \dfrac{11}{4}$
 
You must log in or register to reply here.
Top Bottom