Помоги решить примеры 2 а, б, в, г и 3 а, б, в, г

2. a) \frac{3c+2}{c^2-4c+4} - \frac{5}{c-2} = \frac{3c+2}{(c-2)^2} - \frac{5}{c-2} = \frac{3c+2-5(c-2)}{(c-2)^2} = \frac{3c+2-5c+10}{(c-2)^2} = \frac{-2c+12}{(c-2)^2} = \frac{-2(c-6)}{(c-2)^2} б) \frac{2mn}{m^3+n^3} + \frac{2m}{m^2-n^2} - \frac{1}{m-n} = \frac{2mn}{(m+n)(m^2-mn+n^2)} + \frac{2m}{(m-n)(m+n)} - \frac{1}{m-n} = \frac{2mn+2m(m^2-mn+n^2) - (m+n)(m^2-mn+n^2)}{(m+n)(m^2-mn+n^2)(m-n)} = \frac{2mn+2m^3-2m^2n+2mn^2 - (m^3+n^3)}{(m+n)(m^2-mn+n^2)(m-n)} = \frac{2mn+2m^3-2m^2n+2mn^2 - m^3 - n^3}{(m+n)(m^2-mn+n^2)(m-n)} = \frac{m^3-2m^2n+2mn^2+2mn - n^3}{(m+n)(m^2-mn+n^2)(m-n)} в) \frac{3a(16-3a)}{9a^2-4} + \frac{3(1+2a)}{2-3a} = \frac{3a(16-3a)}{(3a-2)(3a+2)} - \frac{3(1+2a)}{3a-2} = \frac{48a-9a^2 - 3(1+2a)(3a+2)}{(3a-2)(3a+2)} = \frac{48a-9a^2 - 3(3a+2+6a^2+4a)}{(3a-2)(3a+2)} = \frac{48a-9a^2 - 9a-6-18a^2-12a}{(3a-2)(3a+2)} = \frac{-27a^2+27a-6}{(3a-2)(3a+2)} = \frac{-3(9a^2-9a+2)}{(3a-2)(3a+2)} = \frac{-3(3a-2)(3a-1)}{(3a-2)(3a+2)} = \frac{-3(3a-1)}{3a+2} г) \frac{y^2+4}{y^3+8} - \frac{1}{y+2} = \frac{y^2+4}{(y+2)(y^2-2y+4)} - \frac{1}{y+2} = \frac{y^2+4 - (y^2-2y+4)}{(y+2)(y^2-2y+4)} = \frac{y^2+4 - y^2+2y-4}{(y+2)(y^2-2y+4)} = \frac{2y}{(y+2)(y^2-2y+4)} = \frac{2y}{y^3+8} 3. a) \frac{x^2-y^2}{3xy} \cdot \frac{3y}{x-y} = \frac{(x-y)(x+y)3y}{3xy(x-y)} = \frac{x+y}{x} б) \frac{c^2-49}{10cd} : \frac{2c+14}{5d} = \frac{(c-7)(c+7)5d}{10cd(2c+14)} = \frac{(c-7)(c+7)5d}{20cd(c+7)} = \frac{c-7}{4c} в) \frac{x^2-10x+25}{3x+12} : \frac{2x-10}{x^2-16} = \frac{(x-5)^2 (x-4)(x+4)}{3(x+4) 2(x-5)} = \frac{(x-5)(x-4)}{6} г) \frac{t^3+8}{12t^2+27t} : \frac{4t+9}{t^2-2t} = \frac{(t+2)(t^2-2t+4) t(t-2)}{3t(4t+9) (4t+9)} = \frac{(t+2)(t^2-2t+4) (t-2)}{3(4t+9)^2}