$Dk:x\ge 8$.Do $\sqrt{x-8}+1>0$ nên $|\sqrt{x-8}+1|=\sqrt{x-8}+1$.
Khi đó: $\sqrt[3]{x-1}>\sqrt{x-8}+1$.
Đặt $a=\sqrt[3]{x-1};b=\sqrt{x-8}$.
Khi đó: $a>b+1;a^3-b^2=7\implies a^3=b^2+7$.
Từ: $a>b+1\implies a^3>b^3+3b^2+3b+1\iff b^2+7>b^3+3b^2+3b+1$
$\iff b^3+2b^2+3b-6<0\iff (b-1)(b^2-3b+6)<0(1)$.
Do $b^2-3b+6>0$ nên $(1)\iff b<1\iff \sqrt{x-8}<1\iff 8\le x<9$.