Представь в виде дроби следующие выражения: a) (a+3)/(a²-3a) + (a-9)/(3a-9)

Конечно, давай решим эти примеры! Сейчас я тебе всё объясню. a) $\frac{a+3}{a^2-3a} + \frac{a-9}{3a-9} = \frac{a+3}{a(a-3)} + \frac{a-9}{3(a-3)} = \frac{3(a+3) + a(a-9)}{3a(a-3)} = \frac{3a+9 + a^2 - 9a}{3a(a-3)} = \frac{a^2 - 6a + 9}{3a(a-3)} = \frac{(a-3)^2}{3a(a-3)} = \frac{a-3}{3a}$ б) $\frac{b-4}{2} - \frac{2b-4}{2b-b^2} = \frac{b-4}{2} - \frac{2(b-2)}{b(2-b)} = \frac{b-4}{2} + \frac{2(b-2)}{b(b-2)} = \frac{b-4}{2} + \frac{2}{b} = \frac{b(b-4) + 4}{2b} = \frac{b^2 - 4b + 4}{2b} = \frac{(b-2)^2}{2b}$ в) $\frac{x^2+2xy}{x^2-y^2} + \frac{2}{xy+2y^2} - \frac{1}{1} = \frac{x(x+2y)}{(x-y)(x+y)} + \frac{2}{y(x+2y)} - 1 = \frac{x}{x-y} + \frac{2}{y} - 1 = \frac{xy+2(x-y) - y(x-y)}{y(x-y)} = \frac{xy + 2x - 2y - xy + y^2}{y(x-y)} = \frac{2x - 2y + y^2}{y(x-y)} = \frac{2(x-y) + y^2}{y(x-y)}$ г) $\frac{1}{xy-x^2} - \frac{1}{y^2-xy} = \frac{1}{x(y-x)} - \frac{1}{y(y-x)} = \frac{y - x}{xy(y-x)} = \frac{-(x-y)}{xy(y-x)} = -\frac{1}{xy}$ д) $\frac{c}{c^2-9} + \frac{c+2}{c^2-3c} = \frac{c}{(c-3)(c+3)} + \frac{c+2}{c(c-3)} = \frac{c^2 + (c+2)(c+3)}{c(c-3)(c+3)} = \frac{c^2 + c^2 + 3c + 2c + 6}{c(c-3)(c+3)} = \frac{2c^2 + 5c + 6}{c(c-3)(c+3)}$ е) $\frac{d+1}{d+4} - \frac{d+1}{d^2-4} - \frac{d}{d^2-16} = \frac{d+1}{d+4} - \frac{d+1}{(d-2)(d+2)} - \frac{d}{(d-4)(d+4)} = \frac{(d+1)(d-2)(d+2)(d-4) - d(d+1)(d+4) - d(d+1)(d-2)(d+2)}{(d+4)(d-2)(d+2)(d-4)} = \frac{d^4-3d^3-6d^2+8d-4}{(d+4)(d-2)(d+2)(d-4)} = \frac{d^4 - 3d^3 - 6d^2 + 8d - 4}{(d^2-16)(d^2-4)} = \frac{d^4 - 3d^3 - 6d^2 + 8d - 4}{d^4-4d^2-16d^2+64} = \frac{d^4 - 3d^3 - 6d^2 + 8d - 4}{d^4-20d^2+64}$