ПОМОГИТЕ ВО ВЛОЖЕНИИ!!!!!!!!!!!

$\frac{2}{x^2-3x}-\frac{1}{x^2+3x}-\frac{x+1}{x^2-9}=\frac{2}{x(x-3)}-\frac{1}{x(x+3)}-\frac{x+1}{(x-3)(x+3)}=$

$=\frac{2(x+3)-(x-3)-x(x+1)}{x(x-3)(x+3)}=\frac{2x+6-x+3-x^2-x}{x(x-3)(x+3)}=\frac{9-x^2}{x(x-3)(x+3)}=$

$=\frac{-(x^2-9)}{x(x-3)(x+3)}=\frac{-(x-3)(x+3)}{x(x-3)(x+3)}=-\frac{1}{x}$

$\frac{2y+1}{y^2+3y}+\frac{y+2}{3y-y^2}-\frac{1}{y}=\frac{2y+1}{y(y+3)}+\frac{y+2}{y(3-y)}-\frac{1}{y}=$

$\frac{(3-y)(2y+1)+(y+2)(y+3)-(y+3)(3-y)}{y(y+3)(3-y)}=\frac{6y-2y^2+3-y+y^2+2y+3y+6-9+y^2}{y(y+3)(3-y)}$

$=\frac{10y-3}{9y-y^3}$

$\frac{a^2+16a+12}{a^3-8}-\frac{2-3a}{a^2+2a+4}-\frac{3}{a-2}=\frac{a^2+16a+12}{(a-2)(a^2+2a+4)}-\frac{2-3a}{a^2+2a+4}-\frac{3}{a-2}=$

$=\frac{a^2+16a+12-(2-3a)(a-2)-3(a^2+2a+4)}{(a-2)(a^2+2a+4)}=$

$=\frac{a^2+16a+12-2a+3a^2+4-6a-3a^2-6a-12}{(a-2)(a^2+2a+4)}=\frac{a^2+2a+4}{(a-2)(a^2+2a+4)}=\frac{1}{a-2}$

$\frac{2}{4b^2-6b+9}+\frac{4b^2+18}{8b^3+27}-\frac{1}{2b+3}=\frac{2}{4b^2-6b+9}+\frac{4b^2+18}{(2b+3)(4b^2-6b+9)}-\frac{1}{2b+3}=$

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

$=\frac{10b+15}{(2b+3)(4b^2-6b+9)}=\frac{5(2b+3)}{(2b+3)(4b^2-6b+9)}=\frac{5}{4b^2-6b+9}$