toán khó

K

kakashi_hatake

$A=\dfrac{8x^4+19x^2+2}{x^2}=8x^2+19+\dfrac{2}{x^2} \\ 8x^2+\dfrac{2}{x^2} \ge 2.\sqrt{8x^2.\dfrac{2}{x^2}}=8 \rightarrow A \ge 19+8=27$
Dấu đẳng thức xảy ra khi $|x|=\dfrac{1}{\sqrt{2}}$
 
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