[toán 9 ] tìm min

H

hien_vuthithanh

$A= \sqrt{x+4}+|x-2y|$

DK x \geq -4

Ta có $\sqrt{x+4}$ \geq 0 \forall x \geq -4

$|x-2y|$ \geq 0 \forall $x,y$

\Rightarrow A \geq 0

Dấu = \Leftrightarrow $x=-4 ,y=-2$
 
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