Решите неравенство: (3x+1)/(x-3)>0.

Question

\[\frac{3x + 1}{x - 3} > 0\]

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

\[Ответ:\left( - \infty; - \frac{1}{3} \right) \cup (3; + \infty).\]

Accepted Answer

\[\frac{3x + 1}{x - 3} > 0\]

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

\[Ответ:\left( - \infty; - \frac{1}{3} \right) \cup (3; + \infty).\]