$\frac{x^2 - 6x + 5}{x - 5}$
$x^2 - 6x + 5 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * 5 = 36 - 20 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{16}}{2 * 1} = \frac{6 + 4}{2} = \frac{10}{2} = 5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{16}}{2 * 1} = \frac{6 - 4}{2} = \frac{2}{2} = 1$
$x^2 - 6x + 5 = (x - 5)(x - 1)$
$\frac{x^2 - 6x + 5}{x - 5} = \frac{(x - 5)(x - 1)}{x - 5} = x - 1$
Ответ: x − 1

$\frac{2x + 12}{x^2 + 3x - 18}$
$x^2 + 3x - 18 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 1 * (-18) = 9 + 72 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{81}}{2 * 1} = \frac{-3 + 9}{2} = \frac{6}{2} = 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{81}}{2 * 1} = \frac{-3 - 9}{2} = \frac{-12}{2} = -6$
$x^2 + 3x - 18 = (x - 3)(x - (-6)) = (x - 3)(x + 6)$
$\frac{2x + 12}{x^2 + 3x - 18} = \frac{2(x + 6)}{(x - 3)(x + 6)} = \frac{2}{x - 3}$
Ответ: $\frac{2}{x - 3}$

$\frac{x^2 + 9x + 14}{x^2 + 7x}$
$x^2 + 9x + 14 = 0$
$D = b^2 - 4ac = 9^2 - 4 * 1 * 14 = 81 + 56 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-9 + \sqrt{25}}{2 * 1} = \frac{-9 + 5}{2} = \frac{-4}{2} = -2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-9 - \sqrt{25}}{2 * 1} = \frac{-9 - 5}{2} = \frac{-14}{2} = -7$
$x^2 + 9x + 14 = (x - (-2))(x - (-7)) = (x + 2)(x + 7)$
$\frac{x^2 + 9x + 14}{x^2 + 7x} = \frac{(x + 2)(x + 7)}{x(x + 7)} = \frac{x + 2}{x}$
Ответ: $\frac{x + 2}{x}$