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Điều kiện: $x^2\le17$
Đặt: $\sqrt{17-x^2}=y;y\ge0$
Ta nhận được hệ:
$\left\{\begin{array}{l}x+y+xy=9\\x^2+y^2=17\end{array}\right.$
Đặt: $x+y=S;xy=P;S^2\ge4P$
Hệ trở thành: $\left\{\begin{array}{l}S+P=9\\S^2-2P=17\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}P=9-S\\S^2-2(9-S)=17\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}P=9-S\\S^2+2S-35=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}S=5\\P=4\end{array}\right.\\\left\{\begin{array}{l}S=-7\\P=16\end{array}\right.&\rm{loại}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x+y=5\\xy=4\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=4\\y=1\end{array}\right.\\\left\{\begin{array}{l}x=1\\y=4\end{array}\right.\end{array}\right.$
Vậy: $x\in\{1;4\}$
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