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Điều kiện: $\frac{-35}{2}\le x\le\frac{47}{2}$
Đặt: $a=\sqrt[4]{47-2x};b=\sqrt[4]{35+2x}$
Ta nhận được hệ:
$\left\{\begin{array}{l}a+b=4\\a^4+b^4=82\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a=4-b\\(4-b)^4+b^4=82\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a=4-b\\2b^4-16b^3+96b^2-256b+174=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a=4-b\\2(b-3)(b-1)(b^2-4b+29)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}a=1\\b=3\end{array}\right.\\\left\{\begin{array}{l}a=3\\b=1\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=23\\x=-17\end{array}\right.$
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