$\frac{x^2 + 7x + 12}{x^2 + x - 6}$
$x^2 + 7x + 12 = 0$
$D = b^2 - 4ac = 7^2 - 4 * 1 * 12 = 49 - 48 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-7 + \sqrt{1}}{2 * 1} = \frac{-7 + 1}{2} = \frac{-6}{2} = -3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-7 - \sqrt{1}}{2 * 1} = \frac{-7 - 1}{2} = \frac{-8}{2} = -4$
$x^2 + 7x + 12 = (x - (-3))(x - (-4)) = (x + 3)(x + 4)$
$x^2 + x - 6 = 0$
$D = b^2 - 4ac = 1^2 - 4 * 1 * (-6) = 1 + 24 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{25}}{2 * 1} = \frac{-1 + 5}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{25}}{2 * 1} = \frac{-1 - 5}{2} = \frac{-6}{2} = -3$
$x^2 + x - 6 = (x - 2)(x - (-3)) = (x - 2)(x + 3)$
тогда:
$\frac{x^2 + 7x + 12}{x^2 + x - 6} = \frac{(x + 3)(x + 4)}{(x - 2)(x + 3)} = \frac{x + 4}{x - 2}$
Ответ: А) $\frac{x + 4}{x - 2}$