Решите неравенство (3x-1)/(x+8)>=2.

\[\frac{3x - 1}{x + 8} \geq 2^{\backslash x + 8}\]

\[3x - 1 - 2x - 16 = x - 17.\]

\[\frac{x - 17}{x + 8} \geq 0\]

\[x - 17 = 0\]

\[x = 17.\]

\[x + 8 = 0\]

\[x = - 8.\]

\[Ответ:x \in ( - \infty; - 8) \cup \lbrack 17; + \infty).\]