Đặt \(\left\{{}\begin{matrix}\sqrt[2023]{1+x}=a\\\sqrt[2023]{1-x}=b\end{matrix}\right.\) ta được:
\(2a^2+3ab+b^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=-a\\b=-2a\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[2023]{1-x}=-\sqrt[2023]{1+x}\\\sqrt[2023]{1-x}=-2\sqrt[2023]{1+x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=-\left(1+x\right)\left(vn\right)\\1-x=-2^{2023}\left(1+x\right)\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1+2^{2023}}{1-2^{2023}}\)