Đáp án:
$A=8x^2-4x+\dfrac{1}{4x^2}+15$
$=4x^2-4x+1+\dfrac{1}{4x^2}-2.\dfrac{1}{2x}.2x+4x^2+16$
$=(2x-1)^2+(\dfrac{1}{2x}-2x)^2+16$
Ta thấy :$(2x-1)^2≥0∀x$
$(\dfrac{1}{2x}-2x)^2+16≥16∀x$
$⇒(2x-1)^2+(\dfrac{1}{2x}-2x)^2+16≥16∀x$
Dấu bằng xảy ra ⇔$\left \{ {{2x-1=0} \\ {\dfrac{1}{2x}-2x=0}} \right.$
$⇔\left \{ {{x=\dfrac{1}{2}} \\ {\dfrac{1}{2x}=2x}} \right.$
$⇒x= \dfrac{1}{2}$
Vậy Min A=$16⇔x= \dfrac{1}{2}$
