ĐKXĐ: \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-x}=a\\\sqrt{x-1}=b\ge0\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a=1-b\\a^3+b^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=1-a\\a^3+b^2=1\end{matrix}\right.\)
\(\Rightarrow a^3+\left(1-a\right)^2=1\)
\(\Leftrightarrow a^3+a^2-2a=0\)
\(\Leftrightarrow a\left(a^2+a-2\right)=0\Rightarrow\left[{}\begin{matrix}a=0\Rightarrow x=2\\a=1\Rightarrow x=1\\a=-2\Rightarrow x=10\end{matrix}\right.\)
Tổng các nghiệm \(2+1+10=13\)