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viethoang1999

ĐK: $x^2\le 1$
Áp dụng AM-GM ta có:
$x^3\sqrt{1-x^2}=x^2.x\sqrt{1-x^2}\le x^2.\dfrac{x^2+1-x^2}{2}\le 1.\dfrac{1}{2}=\dfrac{1}{2}$
 
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