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

$\frac{3x^2 - 10x + 3}{x - 3} - \frac{x^2 - 4}{x + 2}$
x − 3 ≠ 0
x ≠ 3
и
x + 2 ≠ 0
x ≠ −2
$3x^2 - 10x + 3 = 0$
$D = b^2 - 4ac = (-10)^2 - 4 * 3 * 3 = 100 - 36 = 64 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{10 + \sqrt{64}}{2 * 3} = \frac{10 + 8}{6} = \frac{18}{6} = 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{10 - \sqrt{64}}{2 * 3} = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}$
$3x^2 - 10x + 3 = 3(x - 3)(x - \frac{1}{3}) = (x - 3)(3x - 1)$
тогда:
$\frac{(x - 3)(3x - 1)}{x - 3} - \frac{x^2 - 4}{x + 2} = 3x - 1 - \frac{(x - 2)(x + 2)}{x + 2} = 3x - 1 - (x - 2) = 3x - 1 - x + 2 = 2x + 1$
y = 2x + 1 при x ≠ −2 и x ≠ 3