\(\sqrt[3]{x+3}+\sqrt[3]{5-x}=2\)(\(ĐKXĐ:-3\le x\le5\))
\(\Leftrightarrow x+3+5-x+3\sqrt[3]{\left(x+3\right)^2\left(5-x\right)}+3\sqrt[3]{\left(x+3\right)\left(5-x\right)^2}=8\)
\(\Leftrightarrow3\sqrt[3]{\left(x+3\right)\left(5-x\right)}\left(\sqrt[3]{x+3}+\sqrt[3]{5-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)\left(5-x\right)=0\left(1\right)\\\sqrt[3]{x+3}+\sqrt[3]{5-x}=0\left(2\right)\end{matrix}\right.\)
Giải (1): \(\left(x+3\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)(thỏa mãn đkxđ)
Giải (2): \(\sqrt[3]{x+3}+\sqrt[3]{5-x}=0\)
\(\Leftrightarrow\sqrt[3]{x+3}=-\sqrt[3]{5-x}\)
\(\Leftrightarrow x+3=-5+x\)
\(\Leftrightarrow3=-5\)(vô lý nên loại)
Vậy nghiệm của phương trình trên là \(S=\left\{-3;5\right\}\)
