Đáp án:
\[\left[ \begin{array}{l}
x = 0\\
x = \pm \frac{{\sqrt 5 }}{2}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} = \sqrt[3]{{5x}}\\
\Leftrightarrow {\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}}} \right)^3} = 5x\\
\Leftrightarrow x + 1 + 3.{\sqrt[3]{{x + 1}}^2}.\sqrt[3]{{x - 1}} + 3.\sqrt[3]{{x + 1}}.{\sqrt[3]{{x - 1}}^2} + x - 1 = 5x\\
\Leftrightarrow 3.{\sqrt[3]{{x + 1}}^2}.\sqrt[3]{{x - 1}} + 3.\sqrt[3]{{x + 1}}.{\sqrt[3]{{x - 1}}^2} = 3x\\
\Leftrightarrow \sqrt[3]{{x + 1}}.\sqrt[3]{{x - 1}}.\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}}} \right) = x\\
\Leftrightarrow \sqrt[3]{{x + 1}}.\sqrt[3]{{x - 1}}.\sqrt[3]{{5x}} = x\\
\Leftrightarrow \left( {x + 1} \right).\left( {x - 1} \right).5x = {x^3}\\
\Leftrightarrow \left( {{x^2} - 1} \right).5x = {x^3}\\
\Leftrightarrow 5{x^3} - 5x = {x^3}\\
\Leftrightarrow 4{x^3} - 5x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \pm \frac{{\sqrt 5 }}{2}
\end{array} \right.
\end{array}\)
