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

$\frac{b - 4}{b^3 - b} : (\frac{b - 1}{2b^2 + 3b + 1} - \frac{1}{b^2 - 1})$
$2b^2 + 3b + 1 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 2 * 1 = 9 - 8 = 1 > 0$
$b_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{1}}{2 * 2} = \frac{-3 + 1}{4} = \frac{-2}{4} = -0,5$
$b_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{1}}{2 * 2} = \frac{-3 - 1}{4} = \frac{-4}{4} = -1$
$2b^2 + 3b + 1 = 2(b - (-0,5))(b - (-1)) = 2(b + 0,5)(b + 1) = (2b + 1)(b + 1)$
тогда:
$\frac{b - 4}{b^3 - b} : (\frac{b - 1}{(2b + 1)(b + 1)} - \frac{1}{b^2 - 1}) = \frac{b - 4}{b(b^2 - 1)} : (\frac{b - 1}{(2b + 1)(b + 1)} - \frac{1}{(b - 1)(b + 1)}) = \frac{b - 4}{b(b^2 - 1)} : \frac{(b - 1)^2 - (2b + 1)}{(2b + 1)(b - 1)(b + 1)} = \frac{b - 4}{b(b^2 - 1)} : \frac{b^2 - 2b + 1 - 2b - 1}{(2b + 1)(b^2 - 1)} = \frac{b - 4}{b(b^2 - 1)} : \frac{b^2 - 4b}{(2b + 1)(b^2 - 1)} = \frac{b - 4}{b(b^2 - 1)} : \frac{b(b - 4)}{(2b + 1)(b^2 - 1)} = \frac{b - 4}{b(b^2 - 1)} * \frac{(2b + 1)(b^2 - 1)}{b(b - 4)} = \frac{1}{b} * \frac{2b + 1}{b} = \frac{2b + 1}{b^2}$

$(\frac{c + 2}{c^2 - c - 6} - \frac{2c}{c^2 - 6c + 9}) : \frac{с^2 + 3c}{(2c - 6)^2}$
$c^2 - c - 6 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-6) = 1 + 24 = 25 > 0$
$c_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{25}}{2 * 1} = \frac{1 + 5}{2} = \frac{6}{2} = 3$
$c_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{25}}{2 * 1} = \frac{1 - 5}{2} = \frac{-4}{2} = -2$
$c^2 - c - 6 = (c - 3)(c - (-2)) = (c - 3)(c + 2)$
тогда:
$(\frac{c + 2}{(c - 3)(c + 2)} - \frac{2c}{c^2 - 6c + 9}) : \frac{с^2 + 3c}{(2c - 6)^2} = (\frac{1}{c - 3} - \frac{2c}{(c - 3)^2}) : \frac{с^2 + 3c}{(2c - 6)^2} = \frac{c - 3 - 2c}{(c - 3)^2} : \frac{c(c + 3)}{(2(c - 3))^2} = \frac{-c - 3}{(c - 3)^2} * \frac{4(c - 3)^2}{c(c + 3)} = \frac{-(c + 3)}{1} * \frac{4}{c(c + 3)} = -\frac{4}{c}$

$(\frac{3}{m - 4} + \frac{2m}{m + 1} + \frac{4m - 6}{m^2 - 3m - 4}) * \frac{4m - 16}{2m - 3}$
$m^2 - 3m - 4 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * (-4) = 9 + 16 = 25 > 0$
$m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{25}}{2 * 1} = \frac{3 + 5}{2} = \frac{8}{2} = 4$
$m_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{25}}{2 * 1} = \frac{3 - 5}{2} = \frac{-2}{2} = -1$
$m^2 - 3m - 4 = (m - 4)(m - (-1)) = (m - 4)(m + 1)$
тогда:
$(\frac{3}{m - 4} + \frac{2m}{m + 1} + \frac{4m - 6}{(m - 4)(m + 1)}) * \frac{4m - 16}{2m - 3} = \frac{3(m + 1) + 2m(m - 4) + 4m - 6}{(m - 4)(m + 1)} * \frac{4(m - 4)}{2m - 3} = \frac{3m + 3 + 2m^2 - 8m + 4m - 6}{m + 1} * \frac{4}{2m - 3} = \frac{2m^2 - m - 3}{m + 1} * \frac{4}{2m - 3}$
$2m^2 - m - 3 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 2 * (-3) = 1 + 24 = 25 > 0$
$m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{25}}{2 * 2} = \frac{1 + 5}{4} = \frac{6}{4} = 1,5$
$m_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{25}}{2 * 2} = \frac{1 - 5}{4} = \frac{-4}{4} = -1$
$2m^2 - m - 3 = 2(m - 1,5)(m - (-1)) = (2m - 3)(m + 1)$
тогда:
$\frac{(2m - 3)(m + 1)}{m + 1} * \frac{4}{2m - 3} = 4$