Toán

D

depvazoi

$A=\sqrt{4x^2-6x+10}$
$=\sqrt{(2x-\dfrac{3}{2})^2+\dfrac{31}{4}} \ge \dfrac{\sqrt{31}}{2}$
$(Dấu bằng sảy ra khi x=\dfrac{3}{4})$
$Vậy Min_A=\dfrac{\sqrt{31}}{2}<=>x=\dfrac{3}{4}$
 
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