Đặt $x = t + \frac{1}{2}$$PT \Leftrightarrow {\left( {t + \frac{1}{2}} \right)^4} + {\left( {t - \frac{1}{2}} \right)^4} = \frac{{41}}{8}$
$ \Leftrightarrow {\left[ {{{\left( {t + \frac{1}{2}} \right)}^2} + {{\left( {t - \frac{1}{2}} \right)}^2}} \right]^2} - 2{\left( {t + \frac{1}{2}} \right)^2}{\left( {t - \frac{1}{2}} \right)^2} = \frac{{41}}{8}$
$ \Leftrightarrow {\left( {2{t^2} + \frac{1}{2}} \right)^2} - 2{\left( {{t^2} - \frac{1}{4}} \right)^2} = \frac{{41}}{8}$
$ \Leftrightarrow 4{t^4} + 2{t^2} + \frac{1}{4} - 2\left( {{t^4} - \frac{1}{2}{t^2} + \frac{1}{{16}}} \right) = \frac{{41}}{8}$
$ \Leftrightarrow 2{t^4} + 3{t^2} - 5 = 0$
$\Leftrightarrow \left [ \begin{gathered} {t^2} = 1 \\ {t^2} =- \frac{5}{2} (loại) \end{gathered} \right.$
$\Leftrightarrow \left [ \begin{gathered} t = 1 \\ t = -1 \end{gathered} \right.$
$\Leftrightarrow \left [ \begin{gathered} x = \frac{3}{2} \\ x = -\frac{1}{2} \end{gathered} \right.$