(Toán 9) Tìm GTNN

E

eye_smile

$A=x^2+3x+\dfrac{1}{x}=x^2-x+\dfrac{1}{4}+\dfrac{1}{x}+4x-\dfrac{1}{4}=(x-\dfrac{1}{2})^2+\dfrac{1}{x}+4x-\dfrac{1}{4} \ge 0+2.2-\dfrac{1}{4}=\dfrac{15}{4}$

Dấu = xảy ra \Leftrightarrow $x=\dfrac{1}{2}$
 
You must log in or register to reply here.
Top Bottom