Решить неравенство:

Решите задачу:

$\frac{15}{4+3x-x^2} \ \textgreater \ 1\\\\ \frac{15}{4+3x-x^2} -1\ \textgreater \ 0\\\\ \frac{15-4-3x+x^2}{-(x^2-3x-4)} \ \textgreater \ 0\\\\ \frac{x^2-3x+11}{x^2-3x-4} \ \textless \ 0\\\\x^2-3x-4=0\; \; \Rightarrow \; \; x_1=-1,\; x_2=4\; \; (teorema\; Vieta)\\\\x^2-3x+11=0\; ,\; \; D=9-44=-35\ \textless \ 0\; \; \Rightarrow \; \; x^2-3x+11\ \textgreater \ 0\\\\\Rightarrow \; \; x^2-3x-4\ \textless \ 0\\\\(x+1)(x-4)\ \textless \ 0\quad \quad +++(-1)---(4)+++\\\\x\in (-1,4)$