{ (x+1)/3 < 6
[m]\left\{\begin {matrix}x ≥8-3\\\frac{x+1}[{}-6<0\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x ≥5\\\frac{x+1-3\cdot 6}{3}<0\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x ≥5\\x<17\end {matrix}\right.[/m]
О т в е т.[b][[/b]5;17[b])[/b]