Đáp án:
\[x = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sqrt[3]{{{{\left( {5x + 1} \right)}^2}}} + \sqrt[3]{{{{\left( {5x - 1} \right)}^2}}} + \sqrt[3]{{25{x^2} - 1}} = 1\\
\Leftrightarrow \left( {\sqrt[3]{{5x + 1}} - \sqrt[3]{{5x - 1}}} \right)\left( {\sqrt[3]{{{{\left( {5x + 1} \right)}^2}}} + \sqrt[3]{{{{\left( {5x - 1} \right)}^2}}} + \sqrt[3]{{25{x^2} - 1}}} \right) = \left( {\sqrt[3]{{5x + 1}} - \sqrt[3]{{5x - 1}}} \right)\\
\Leftrightarrow \left( {5x + 1} \right) - \left( {5x - 1} \right) = \left( {\sqrt[3]{{5x + 1}} - \sqrt[3]{{5x - 1}}} \right)\\
\Leftrightarrow 2 = \left( {\sqrt[3]{{5x + 1}} - \sqrt[3]{{5x - 1}}} \right)\\
\Leftrightarrow {\left( {\sqrt[3]{{5x + 1}} - \sqrt[3]{{5x - 1}}} \right)^3} = 8\\
\Leftrightarrow \left( {5x + 1} \right) - \left( {5x - 1} \right) - 3.\sqrt[3]{{\left( {5x - 1} \right)\left( {5x + 1} \right)}}.\left( {\sqrt[3]{{5x + 1}} - \sqrt[3]{{5x - 1}}} \right) = 8\\
\Leftrightarrow 2 - 3.\sqrt[3]{{\left( {5x - 1} \right)\left( {5x + 1} \right)}}.2 = 8\\
\Leftrightarrow \sqrt[3]{{\left( {5x - 1} \right)\left( {5x + 1} \right)}} = - 1\\
\Leftrightarrow \left( {5x - 1} \right)\left( {5x + 1} \right) = - 1\\
\Leftrightarrow 25{x^2} - 1 = - 1\\
\Leftrightarrow 25{x^2} = 0\\
\Leftrightarrow x = 0
\end{array}\)
