1. ĐK: $x\ge0$
Đặt: $u=\sqrt{x};v=\sqrt[4]{17-x^2};u,v\ge0$
Khi đó, ta có hệ phương trình:
$\left\{\begin{array}{l}u+v=3\\u^4+v^4=17\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}v=3-u\\u^4+(3-u)^4=17\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}v=3-u\\2(u-1)(u-2)(u^2-3u+16)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}u=1\\v=2\end{array}\right.\\\left\{\begin{array}{l}u=2\\v=1\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=1\\x=4\end{array}\right.$