Вопрос:

Решите неравенство: (x^2+x-12)/(x^2-x-2)>=0.

Ответ:

\[\frac{x^{2} + x - 12}{x^{2} - x - 2} \geq 0\]

\[x^{2} + x - 12 =\]

\[= x^{2} + 4x - 3x - 12 =\]

\[= x(x + 4) - 3(x + 4) =\]

\[= (x + 4)(x - 3)\]

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

\[= x(x - 2) + (x - 2) =\]

\[= (x - 2)(x + 1)\]

\[\frac{(x + 4)(x - 3)}{(x + 1)(x - 2)} \geq 0\]

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