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

$\frac{x - 4}{x^2 - 10x + 24}$
$x^2 - 10x + 24 = 0$
$D = b^2 - 4ac = (-10)^2 - 4 * 1 * 24 = 100 + 96 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{10 + \sqrt{4}}{2 * 1} = \frac{10 + 2}{2} = \frac{12}{2} = 6$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{10 - \sqrt{4}}{2 * 1} = \frac{10 - 2}{2} = \frac{8}{2} = 4$
$x^2 - 10x + 24 = (x - 6)(x - 4)$
$\frac{x - 4}{x^2 - 10x + 24} = \frac{x - 4}{(x - 6)(x - 4)} = \frac{1}{x - 6}$
Ответ: $\frac{1}{x - 6}$

$\frac{3x - 15}{x^2 - x - 20}$
$x^2 - x - 20 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-20) = 1 + 80 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{81}}{2 * 1} = \frac{1 + 9}{2} = \frac{10}{2} = 5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{81}}{2 * 1} = \frac{1 - 9}{2} = \frac{-8}{2} = -4$
$x^2 - x - 20 = (x - 5)(x - (-4)) = (x - 5)(x + 4)$
$\frac{3x - 15}{x^2 - x - 20} = \frac{3(x - 5)}{(x - 5)(x + 4)} = \frac{3}{x + 4}$
Ответ: $\frac{3}{x + 4}$

$\frac{x^2 - 3x + 2}{6x - 6}$
$x^2 - 3x + 2 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * 2 = 9 - 8 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{1}}{2 * 1} = \frac{3 + 1}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{1}}{2 * 1} = \frac{3 - 1}{2} = \frac{2}{2} = 1$
$x^2 - 3x + 2 = (x - 2)(x - 1)$
$\frac{x^2 - 3x + 2}{6x - 6} = \frac{(x - 2)(x - 1)}{6(x - 1)} = \frac{x - 2}{6}$
Ответ: $\frac{x - 2}{6}$

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

$\frac{x^2 + 4x}{x^2 + 2x - 8}$
$x^2 + 2x - 8 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-8) = 4 + 32 = 36 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{36}}{2 * 1} = \frac{-2 + 6}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{36}}{2 * 1} = \frac{-2 - 6}{2} = \frac{-8}{2} = -4$
$x^2 + 2x - 8 = (x - 2)(x - (-4)) = (x - 2)(x + 4)$
$\frac{x^2 + 4x}{x^2 + 2x - 8} = \frac{x(x + 4)}{(x - 2)(x + 4)} = \frac{x}{x - 2}$
Ответ: $\frac{x}{x - 2}$