ĐK: $x^8\le17$
Đặt: $\sqrt[4]{17-x^8}=u;\sqrt[3]{2x^8-1}=v;u\ge0$
Ta có hệ phương trình:
$\left\{\begin{array}{l}u-v=1\\2u^4+v^3=33\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}v=u-1\\2u^4+(u-1)^3=33\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}v=u-1\\2u^4+u^3-3u^2+3u-34=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}v=u-1\\(u-2)(2u^3+5u^2+7u+17)=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}u=2\\v=1\end{array}\right.$
Khi đó ta có:
$\left\{\begin{array}{l}\sqrt[4]{17-x^8}=2\\\sqrt[3]{2x^8-1}=1\end{array}\right.$
$\Leftrightarrow x^8=1$
$\Leftrightarrow x=\pm1$