Решите неравенство: (x^2+4x+4)/(x^2-9x+8)<=0.

\[\frac{x^{2} + 4x + 4}{x^{2} - 9x + 8} \leq 0\]

\[x^{2} - 9x + 8 =\]

\[= x^{2} - 8x - x + 8 =\]

\[= x(x - 8) - (x - 8) =\]

\[= (x - 8)(x - 1)\]

\[\frac{(x + 2)^{2}}{(x - 8)(x - 1)} \leq 0\]

\[Ответ:\left\{ - 2 \right\} \cup (1;8).\]