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Điều kiện: $-5\le x\le 13$
Đặt: $\sqrt{x+5}=a;\sqrt{13-x}=b;a;b\ge0$
Ta nhận được hệ:
$\left\{\begin{array}{l}a-ab+b=-3\\a^2+b^2=18\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=ab-3\\(a+b)^2-2ab=18\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=ab-3\\(ab-3)^2-2ab-18=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=ab-3\\a^2b^2-8ab-9=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a+b=6\\ab=9\end{array}\right.$ (vì $a;b\ge0$)
$\Leftrightarrow \left\{\begin{array}{l}a=3\\b=3\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\sqrt{x+5}=3\\\sqrt{13-x}=3\end{array}\right.\Leftrightarrow x=4$
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