ĐKXĐ: \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-x}=a\le1\\\sqrt{x-1}=b\ge0\end{matrix}\right.\) ta được hệ pt:
\(\left\{{}\begin{matrix}a+b-1=0\\a^3+b^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=1-a\\a^3+b^2-1=0\end{matrix}\right.\)
\(\Rightarrow a^3+\left(1-a\right)^2-1=0\Rightarrow a^3+a^2-2a=0\Rightarrow a\left(a^2+a-2\right)=0\)
\(\Rightarrow a\left(a-1\right)\left(a+2\right)=0\) \(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{2-x}=0\\\sqrt[3]{2-x}=1\\\sqrt[3]{2-x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2-x=0\\2-x=1\\2-x=-8\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=10\end{matrix}\right.\)
