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

$\frac{2y^2 + 3y - 5}{y^2 - 2y + 1}$
$2y^2 + 3y - 5 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 2 * (-5) = 9 + 40 = 49 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{49}}{2 * 2} = \frac{-3 + 7}{4} = \frac{4}{4} = 1$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{49}}{2 * 2} = \frac{-3 - 7}{4} = \frac{-10}{4} = -2,5$
$2y^2 + 3y - 5 = 2(y - 1)(y - (-2,5)) = 2(y - 1)(y + 2,5) = (y - 1)(2y + 5)$
$\frac{2y^2 + 3y - 5}{y^2 - 2y + 1} = \frac{(y - 1)(2y + 5)}{(y - 1)^2} = \frac{2y + 5}{y - 1}$
Ответ: $\frac{2y + 5}{y - 1}$

$\frac{a^2 + 5a + 4}{a^2 - a - 20}$
$a^2 + 5a + 4 = 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * 4 = 25 - 16 = 9 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{9}}{2 * 1} = \frac{-5 + 3}{2} = \frac{-2}{2} = -1$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{9}}{2 * 1} = \frac{-5 - 3}{2} = \frac{-8}{2} = -4$
$a^2 + 5a + 4 = (a - (-1))(a - (-4)) = (a + 1)(a + 4)$
$a^2 - a - 20 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-20) = 1 + 80 = 81 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{81}}{2 * 1} = \frac{1 + 9}{2} = \frac{10}{2} = 5$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{81}}{2 * 1} = \frac{1 - 9}{2} = \frac{-8}{2} = -4$
$a^2 - a - 20 = (a - 5)(a - (-4)) = (a - 5)(a + 4)$
$\frac{a^2 + 5a + 4}{a^2 - a - 20} = \frac{(a + 1)(a + 4)}{(a - 5)(a + 4)} = \frac{a + 1}{a - 5}$
Ответ: $\frac{a + 1}{a - 5}$

$\frac{3 + 20b - 7b^2}{7b^2 - 6b - 1}$
$3 + 20b - 7b^2 = -7b^2 + 20b + 3 = 0$
$D = b^2 - 4ac = 20^2 - 4 * (-7) * 3 = 400 + 84 = 484 > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-20 + \sqrt{484}}{2 * (-7)} = \frac{-20 + 22}{-14} = \frac{2}{-14} = -\frac{1}{7}$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-20 - \sqrt{484}}{2 * (-7)} = \frac{-20 - 22}{-14} = \frac{-42}{-14} = 3$
$-7b^2 + 20b + 3 = -7(b - (-\frac{1}{7}))(b - 3) = -7(b + \frac{1}{7})(b - 3) = (7b + 1)(3 - b)$
$7b^2 - 6b - 1 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 7 * (-1) = 36 + 28 = 64 > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{64}}{2 * 7} = \frac{6 + 8}{14} = \frac{14}{14} = 1$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{64}}{2 * 7} = \frac{6 - 8}{14} = \frac{-2}{14} = -\frac{1}{7}$
$7b^2 - 6b - 1 = 7(b - 1)(b - (-\frac{1}{7})) = 7(b - 1)(b + \frac{1}{7}) = (b - 1)(7b + 1)$
$\frac{3 + 20b - 7b^2}{7b^2 - 6b - 1} = \frac{(7b + 1)(3 - b)}{(b - 1)(7b + 1)} = \frac{3 - b}{b - 1}$
Ответ: $\frac{3 - b}{b - 1}$