Решите неравенство: 64x^2-16x+1<=0.

\[64x^{2} - 16x + 1 \leq 0\]

\[(8x - 1)^{2} \leq 0\]

\[x = \frac{1}{8}\]

\[Ответ:\ \ x \in \left\{ \frac{1}{8} \right\}\text{.\ }\]