Решите неравенство: x-(3x-1)/3+(x+1)/2>=1.

\[x - \frac{3x - 1}{3} + \frac{x + 1}{2} \geq 1\ \ \ \ \ \ | \cdot 6\]

\[6x - 2 \cdot (3x - 1) + 3 \cdot (x + 1) \geq 6\]

\[6x - 6x + 2 + 3x + 3 \geq 6\]

\[3x \geq 1\]

\[x \geq \frac{1}{3}\]

\[Ответ:x \in \left\lbrack \frac{1}{3}; + \infty \right).\]