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Đặt: $\sqrt[3]{8-x}=a,\sqrt[3]{x+27}=b$
Ta nhận được hệ:
$\left\{\begin{array}{l}a^2-ab+b^2=7\\a^3+b^3=35\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=5\\a^2-ab+b^2=7\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=5\\(a+b)^2-3ab=7\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=5\\ab=6\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}a=2\\b=3\end{array}\right.\\\left\{\begin{array}{l}a=3\\b=2\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=-19\\x=0\end{array}\right.$
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