Решите неравенство: (x^2+3x-4)/(x-1)>0.

Ответ:

\[\frac{x^{2} + 3x - 4}{x - 1} > 0\ \]

\[x^{2} + 3x - 4 = (x + 4)(x - 1)\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = - 4;\ \ \ x_{2} = 1.\]

\[\frac{(x + 4)(x - 1)}{x - 1} > 0;\ \ x \neq 1.\]

\[x + 4 > 0\]

\[x > - 4;\ \ x \neq 1.\]

\[x \in ( - 4;1) \cup (1; + \infty).\]