Đáp án:
\[\left[ \begin{array}{l}
x = 0\\
x = - 8
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {x + \frac{1}{x}} \right)^2} = {x^2} + 2.x.\frac{1}{x} + \frac{1}{{{x^2}}} = \left( {{x^2} + \frac{1}{{{x^2}}}} \right) + 2\\
8{\left( {x + \frac{1}{x}} \right)^2} + 4{\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} - 4.\left( {{x^2} + \frac{1}{{{x^2}}}} \right){\left( {x + \frac{1}{x}} \right)^2} = {\left( {x + 4} \right)^2}\\
\Leftrightarrow 8{\left( {x + \frac{1}{x}} \right)^2} + 4.{\left[ {{{\left( {x + \frac{1}{x}} \right)}^2} - 2} \right]^2} - 4.\left[ {{{\left( {x + \frac{1}{x}} \right)}^2} - 2} \right].{\left( {x + \frac{1}{x}} \right)^2} = {\left( {x + 4} \right)^2}\\
\Leftrightarrow 8{\left( {x + \frac{1}{x}} \right)^2} + 4.{\left( {x + \frac{1}{x}} \right)^4} - 16{\left( {x + \frac{1}{x}} \right)^2} + 16 - 4.{\left( {x + \frac{1}{x}} \right)^4} + 8.{\left( {x + \frac{1}{x}} \right)^2} = {\left( {x + 4} \right)^2}\\
\Leftrightarrow 16 = {\left( {x + 4} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x + 4 = 4\\
x + 4 = - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 8
\end{array} \right.
\end{array}\)
