Решите уравнение 121x^3-22x^2+x=0.

\[121x^{3} - 22x^{2} + x = 0\]

\[x\left( 121x^{2} - 22x + 1 \right) = 0\]

\[x(11x - 1)^{2} = 0\]

\[x = 0;\ \ \ \ \ \ \ \ 11x - 1 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 11x = 1\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{1}{11}\]

\[Ответ:\ \ x = 0;\ \ x = \frac{1}{11}.\]