\[16x^{2} + \frac{1}{x^{2}} = 89;\ \ \ 4x - \frac{1}{x} = ?\]
\[16x^{2} - 8x \cdot \frac{1}{x} + \frac{1}{x^{2}} = 89 - 8x \cdot \frac{1}{x}\]
\[\left( 4x - \frac{1}{x} \right)^{2} = 89 - 8\]
\[\left( 4x - \frac{1}{x} \right)^{2} = 81\]
\[4x - \frac{1}{x} = 9;\]
\[4x - \frac{1}{x} = - 9.\]
\[Ответ:\ - 9;\ \ 9.\]