Решите неравенство: 3x^2-4x+1<0.

\[3x^{2} - 4x + 1 < 0\]

\[D_{1} = 4 - 3 = 1\]

\[x_{1} = \frac{2 + 1}{3} = 1;\]

\[x_{2} = \frac{2 - 1}{3} = \frac{1}{3};\]

\[\left( x - \frac{1}{3} \right)(x - 1) < 0\]

\[\frac{1}{3} < x < 1.\]

\[Ответ:\ \frac{1}{3} < x < 1.\]