Đáp án:
$\begin{array}{l}
{x^2} - 5x + 1 = 0\\
Do\,pt\,có\,nghiệm\,x \ne 0\\
\Rightarrow {x^2}\left( {{x^2} - 5x + 1} \right) = 0\\
\Rightarrow {x^4} - 5{x^2} + {x^2} = 0\\
\Rightarrow {x^4} - 5{x^2} + 1 = 1 - {x^2}\\
\Rightarrow M = \frac{{{x^4} - 5{x^2} + 1}}{{8{x^2}}} = \frac{{1 - {x^2}}}{{8{x^2}}} = \frac{1}{{8{x^2}}} - \frac{1}{8}
\end{array}$