$\frac{4a^2 - 9}{2a^2 - 9a - 18}$
$2a^2 - 9a - 18 = 0$
$D = b^2 - 4ac = (-9)^2 - 4 * 2 * (-18) = 81 + 144 = 251 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{251}}{2 * 2} = \frac{9 + 15}{4} = \frac{24}{4} = 6$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{251}}{2 * 2} = \frac{9 - 15}{4} = \frac{-6}{4} = -1,5$
$2a^2 - 9a - 18 = 2(a - 6)(a - (-1,5)) = (a - 6)(2a + 3)$
$\frac{4a^2 - 9}{2a^2 - 9a - 18} = \frac{(2a - 3)(2a + 3)}{(a - 6)(2a + 3)} = \frac{2a - 3}{a - 6}$
Ответ: $\frac{2a - 3}{a - 6}$

$\frac{2b^2 - 7b + 3}{4b^2 - 4b + 1}$
$2b^2 - 7b + 3 = 0$
$D = b^2 - 4ac = (-7)^2 - 4 * 2 * 3 = 49 + 24 = 25 > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{25}}{2 * 2} = \frac{7 + 5}{4} = \frac{12}{4} = 3$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{25}}{2 * 2} = \frac{7 - 5}{4} = \frac{2}{4} = 0,5$
$2b^2 - 7b + 3 = 2(b - 3)(b - 0,5) = (b - 3)(2b - 1)$
$\frac{2b^2 - 7b + 3}{4b^2 - 4b + 1} = \frac{(b - 3)(2b - 1)}{(2b - 1)^2} = \frac{b - 3}{2b - 1}$
Ответ: $\frac{b - 3}{2b - 1}$

$\frac{c^2 - 5c - 6}{c^2 - 8c + 12}$
$c^2 - 5c - 6 = 0$
$D = b^2 - 4ac = (-5)^2 - 4 * 1 * (-6) = 25 + 24 = 49 > 0$
$c_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{49}}{2 * 1} = \frac{5 + 7}{2} = \frac{12}{2} = 6$
$c_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{49}}{2 * 1} = \frac{5 - 7}{2} = \frac{-2}{2} = -1$
$c^2 - 5c - 6 = (c - 6)(c - (-1)) = (c - 6)(c + 1)$
$c^2 - 8c + 12 = 0$
$D = b^2 - 4ac = (-8)^2 - 4 * 1 * 12 = 64 + 48 = 16 > 0$
$c_1 = \frac{-b + \sqrt{D}}{2a} = \frac{8 + \sqrt{16}}{2 * 1} = \frac{8 + 4}{2} = \frac{12}{2} = 6$
$c_2 = \frac{-b - \sqrt{D}}{2a} = \frac{8 - \sqrt{16}}{2 * 1} = \frac{8 - 4}{2} = \frac{4}{2} = 2$
$c^2 - 8c + 12 = (c - 6)(c - 2)$
$\frac{c^2 - 5c - 6}{c^2 - 8c + 12} = \frac{(c - 6)(c + 1)}{(c - 6)(c - 2)} = \frac{c + 1}{c - 2}$
Ответ: $\frac{c + 1}{c - 2}$

$\frac{m^3 - 1}{m^2 + 9m - 10}$
$m^2 + 9m - 10 = 0$
$D = b^2 - 4ac = 9^2 - 4 * 1 * (-10) = 81 + 40 = 121 > 0$
$m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-9 + \sqrt{121}}{2 * 1} = \frac{-9 + 11}{2} = \frac{2}{2} = 1$
$m_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-9 - \sqrt{121}}{2 * 1} = \frac{-9 - 11}{2} = \frac{-20}{2} = -10$
$m^2 + 9m - 10 = (m - 1)(m - (-10)) = (m - 1)(m + 10)$
$\frac{m^3 - 1}{m^2 + 9m - 10} = \frac{(m - 1)(m^2 + m + 1)}{(m - 1)(m + 10)} = \frac{m^2 + m + 1}{m + 10}$
Ответ: $\frac{m^2 + m + 1}{m + 10}$

$\frac{x^2 - 16}{32 - 4x - x^2}$
$32 - 4x - x^2 = -x^2 - 4x + 32 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * (-1) * 32 = 16 + 128 = 144 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{144}}{2 * (-1)} = \frac{4 + 12}{-2} = \frac{16}{-2} = -8$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{144}}{2 * (-1)} = \frac{4 - 12}{-2} = \frac{-8}{-2} = 4$
$-x^2 - 4x + 32 = -(x - (-8))(x - 4) = -(x + 8)(x - 4)$
$\frac{x^2 - 16}{32 - 4x - x^2} = -\frac{(x - 4)(x + 4)}{(x + 8)(x - 4)} = -\frac{x + 4}{x + 8}$
Ответ: $-\frac{x + 4}{x + 8}$

$\frac{4n^2 - 9n + 2}{2 + 9n - 5n^2}$
$4n^2 - 9n + 2 = 0$
$D = b^2 - 4ac = (-9)^2 - 4 * 4 * 2 = 81 - 32 = 49 > 0$
$n_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{49}}{2 * 4} = \frac{9 + 7}{8} = \frac{16}{8} = 2$
$n_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{49}}{2 * 4} = \frac{9 - 7}{8} = \frac{2}{8} = \frac{1}{4}$
$4n^2 - 9n + 2 = 4(n - 2)(n - \frac{1}{4}) = (n - 2)(4n - 1)$
$2 + 9n - 5n^2 = -5n^2 + 9n + 2 = 0$
$D = b^2 - 4ac = 9^2 - 4 * (-5) * 2 = 81 + 40 = 121 > 0$
$n_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-9 + \sqrt{121}}{2 * (-5)} = \frac{-9 + 11}{-10} = \frac{2}{-10} = -0,2$
$n_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-9 - \sqrt{121}}{2 * (-5)} = \frac{-9 - 11}{-10} = \frac{-20}{-10} = 2$
$-5n^2 + 9n + 2 = -5(n - (-0,2))(n - 2) = -5(n + 0,2)(n - 2) = -(5n + 1)(n - 2)$
$\frac{4n^2 - 9n + 2}{2 + 9n - 5n^2} = \frac{(n - 2)(4n - 1)}{-(5n + 1)(n - 2)} = -\frac{4n - 1}{5n + 1} = \frac{1 - 4n}{5n + 1}$
Ответ: $\frac{1 - 4n}{5n + 1}$