{Đại 9} Tìm min

D

depvazoi

$A=\dfrac{x-1}{x+2}=1-\dfrac{3}{x+2}$
Vì $x \ge 0 => 1-\dfrac{3}{x+2} \ge -\dfrac{1}{2}$ (Dấu bằng xảy ra khi x=0)
Vậy $Min_A=-\dfrac{1}{2}<=>x=0$
 
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