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Điều kiện: $|x|\le 1$
Đặt: $a=\sqrt{1-x^2};b=\sqrt[3]{1-x^2}$
Ta nhận được hệ:
$\left\{\begin{array}{l}a+2b=3\\a^2+b^3=2\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a=3-2b\\(3-2b)^2+b^3=2\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a=3-2b\\b^3+4b^2-12b+7=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}a=1\\b=1\end{array}\right.\\\left\{\begin{array}{l}a=8+\sqrt{33}\\b=\frac{1}{2}(-5-\sqrt{33})\end{array}\right.\\\left\{\begin{array}{l}a=8-\sqrt{33}\\b=\frac{1}{2}(-5+\sqrt{33})\end{array}\right.\end{array}\right.$
$\Leftrightarrow x=0$
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