910. Упростите: a) $$\frac{2}{x^2-3x}-\frac{1}{x^2+3x}-\frac{x+1}{x^2-9}$$; б) $$\frac{2y+1}{y^2+3y}+\frac{y+2}{3y-y^2}$$.

a) $$\frac{2}{x^2-3x}-\frac{1}{x^2+3x}-\frac{x+1}{x^2-9}=\frac{2}{x(x-3)}-\frac{1}{x(x+3)}-\frac{x+1}{(x-3)(x+3)}=\frac{2(x+3)-1(x-3)-x(x+1)}{x(x-3)(x+3)}=\frac{2x+6-x+3-x^2-x}{x(x-3)(x+3)}=\frac{-x^2+9}{x(x-3)(x+3)}=\frac{-(x^2-9)}{x(x-3)(x+3)}=\frac{-(x-3)(x+3)}{x(x-3)(x+3)}=-\frac{1}{x}$$
б) $$\frac{2y+1}{y^2+3y}+\frac{y+2}{3y-y^2}=\frac{2y+1}{y(y+3)}+\frac{y+2}{-y(y-3)}=\frac{2y+1}{y(y+3)}-\frac{y+2}{y(y-3)}=\frac{(2y+1)(y-3)-(y+2)(y+3)}{y(y+3)(y-3)}=\frac{2y^2-6y+y-3-(y^2+3y+2y+6)}{y(y+3)(y-3)}=\frac{2y^2-5y-3-y^2-5y-6}{y(y+3)(y-3)}=\frac{y^2-10y-9}{y(y+3)(y-3)}$$