Đặt \(\sqrt[3]{2-x}=a\) \(\Rightarrow\left\{{}\begin{matrix}1-x=a^3-1\\x=2-a^3\end{matrix}\right.\) ( \(a\ge1\); \(x\le1\) )
Pt \(\Leftrightarrow\left(1-\sqrt{a^3-1}\right)\cdot a=2-a^3\)
\(\Leftrightarrow a^3+a-2-a\sqrt{a^3-1}=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+a+2\right)-a\sqrt{a-1}\cdot\sqrt{a^2+a+1}=0\)
\(\Leftrightarrow\sqrt{a-1}\left[\sqrt{a-1}\left(a^2+a+2\right)-a\sqrt{a^2+a+1}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a-1}=0\\\sqrt{a-1}\left(a^2+a+2\right)=a\sqrt{a^2+a+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\\left(a-1\right)\left(a^2+a+2\right)^2=a^2\left(a^2+a+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{2-x}=1\\a^5+2a^3-2a^2-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=1\\\left(a^3-2\right)\left(a^2+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\a^3=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2-x=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\) ( thỏa )
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