Đặt \(\sqrt[3]{x}=a\); \(\sqrt[3]{1-x}=b\)
Khi đó: \(a^3+b^3=x+1-x=1\)
Ta có hệ \(\left\{{}\begin{matrix}a+b=1\\a^3+b^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1-b\left(1\right)\\\left(1-b\right)^3+b^3=1\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow3b^2-3b=0\)
\(\Leftrightarrow3b\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=0\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\\\left\{{}\begin{matrix}a=0\\b=1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt[3]{x}=1\\\sqrt[3]{1-x}=0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt[3]{x}=0\\\sqrt[3]{1-x}=1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\x=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\x=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
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