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Điều kiện: $x\le \frac{1}{2}$
Đặt: $a=\sqrt{\frac{1}{2}-x};b=\sqrt[3]{\frac{1}{2}+x}$
Ta nhận được hệ:
$\left\{\begin{array}{l}a+b=1\\a^2+b^3=1\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a=1-b\\(1-b)^2+b^3=1\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a=1-b\\b^3+b^2-2b=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}a=1\\b=0\end{array}\right.\\\left\{\begin{array}{l}a=0\\b=1\end{array}\right.\\\left\{\begin{array}{l}a=3\\b=-2\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\frac{1}{2}\\x=-\frac{1}{2}\\x=-\frac{-17}{2}\end{array}\right.$
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