Question

Помогите пожалуйста ! Номер 1 и 2 . Даю 70 баллов

Answer 1

1. а)6х/ 2х^2-6х = 6х/2х(х-3)=3/(х-3)

б) 12х+12у/36х^2-36у^2=12(х+у)/12(3х^2-3у^2)= х+у/3(х^2-у^2)= х+у/3(х+у)(х-у)=1/3(х-у)

Answer 2

№ 1

а)

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

б)

$\frac{12x+12y}{36x^{2}-36y^{2}}=\frac{12x+12y}{(6x)^{2}-(6y)^{2}}=\frac{2(6x+6y)}{(6x-6y)(6x+6y)}=\frac{2}{6x-6y}=\frac{1}{3(x-y)}$

№ 2

a)

$\frac{3}{14x-14y}-\frac{2}{21x-21y}=\frac{3}{2(7x-7y)}-\frac{2}{3(7x-7y)}=\frac{9}{6(7x-7y)}-\frac{4}{6(7x-7y)}=\frac{5}{6(7x-7y)}=\frac{5}{42(x-y)}$

б)

$\frac{2-3y}{y^{2}-9}+\frac{5-2y}{y-3}=\frac{2-3y}{(y-3)(y+3)}+\frac{5-2y}{y-3}=\frac{2-3y}{(y-3)(y+3)}+\frac{(5-2y)(y+3)}{(y-3)(y+3)}=\\ \\=\frac{2-3y+(5-2y)(y+3)}{(y-3)(y+3)}=\frac{2-3y+5y+15-2y^{2}-6y}{(y-3)(y+3)}=\frac{-2y^{2}-4y+17}{(y-3)(y+3)}$