Question

ПОМОГИТЕ ВО ВЛОЖЕНИИ!!!!!!!!!!!

Answer 1

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

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

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

 

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

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

$=frac{10y-3}{9y-y^3}$ 

 

 

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

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

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

 

 

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

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

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