giải pt ∛(5x+1)^2 + ∛(5x-1)^2 + ∛25x^2-1 =1

Đáp án:

\[x = 0\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\sqrt[3]{{{{\left( {5x + 1} \right)}^2}}} + \sqrt[3]{{{{\left( {5x – 1} \right)}^2}}} + \sqrt[3]{{25{x^2} – 1}} = 1\\
 \Leftrightarrow \left( {\sqrt[3]{{5x + 1}} – \sqrt[3]{{5x – 1}}} \right)\left( {\sqrt[3]{{{{\left( {5x + 1} \right)}^2}}} + \sqrt[3]{{{{\left( {5x – 1} \right)}^2}}} + \sqrt[3]{{25{x^2} – 1}}} \right) = \left( {\sqrt[3]{{5x + 1}} – \sqrt[3]{{5x – 1}}} \right)\\
 \Leftrightarrow \left( {5x + 1} \right) – \left( {5x – 1} \right) = \left( {\sqrt[3]{{5x + 1}} – \sqrt[3]{{5x – 1}}} \right)\\
 \Leftrightarrow 2 = \left( {\sqrt[3]{{5x + 1}} – \sqrt[3]{{5x – 1}}} \right)\\
 \Leftrightarrow {\left( {\sqrt[3]{{5x + 1}} – \sqrt[3]{{5x – 1}}} \right)^3} = 8\\
 \Leftrightarrow \left( {5x + 1} \right) – \left( {5x – 1} \right) – 3.\sqrt[3]{{\left( {5x – 1} \right)\left( {5x + 1} \right)}}.\left( {\sqrt[3]{{5x + 1}} – \sqrt[3]{{5x – 1}}} \right) = 8\\
 \Leftrightarrow 2 – 3.\sqrt[3]{{\left( {5x – 1} \right)\left( {5x + 1} \right)}}.2 = 8\\
 \Leftrightarrow \sqrt[3]{{\left( {5x – 1} \right)\left( {5x + 1} \right)}} =  – 1\\
 \Leftrightarrow \left( {5x – 1} \right)\left( {5x + 1} \right) =  – 1\\
 \Leftrightarrow 25{x^2} – 1 =  – 1\\
 \Leftrightarrow 25{x^2} = 0\\
 \Leftrightarrow x = 0
\end{array}\)