ПОМОГИТЕ!!!! СРОЧНО!!!!Сократите дроби: 1) 3y+9/y^2+y-6 2)x^2+6x+5/x^2+3x+2...

Решите задачу:

$\frac{3y+9}{y^2+y-6} = \frac{3(y+3)}{y^2+3y-2y-6} = \frac{3(y+3)}{y(y+3) -2(y+3)} = \frac{3(y+3)}{(y-2)(y+3)} = \frac{3}{y-2}$

$\frac{x^2+6x+5}{x^2+3x+2} = \frac{x^2+5x+x+5}{x^2+2x+x+2} = \frac{x(x+5) +1(x+5)}{x(x+2) +1(x+2)} = \frac{(x+5)(x+1)}{(x+2)(x+1)} = \frac{x+5}{x+2}$

$\frac{2x^3-9x^2 +27x-27}{4x^2-9} = \frac{2x^3-6x^2-3x^2+9x+18x-27}{(2x)^2-3^2} = \\ \\ = \frac{(2x^3-6x^2+18x) + (-3x^2+9x-27)}{(2x-3)(2x+3)} = \\ \\ = \frac{2x(x^2-x+9) -3(x^2-3x+9)}{(2x-3)(2x+3)} = \\ \\ = \frac{(2x-3)(x^2-3x+9)}{(2x-3)(2x+3)} = \frac{x^2-3x+9}{2x+3}$