Помоги сократить дробь в задании 32

Задание 32. Сократить дробь: a) $\frac{y^2-16}{3y+12}$ = $\frac{(y-4)(y+4)}{3(y+4)}$ = $\frac{y-4}{3}$ б) $\frac{5x-15y}{x^2-9y^2}$ = $\frac{5(x-3y)}{(x-3y)(x+3y)}$ = $\frac{5}{x+3y}$ в) $\frac{(c+2)^2}{7c^2+14c}$ = $\frac{(c+2)(c+2)}{7c(c+2)}$ = $\frac{c+2}{7c}$ г) $\frac{6cd-18c}{(d-3)^2}$ = $\frac{6c(d-3)}{(d-3)(d-3)}$ = $\frac{6c}{d-3}$ д) $\frac{a^2+10a+25}{a^2-25}$ = $\frac{(a+5)(a+5)}{(a-5)(a+5)}$ = $\frac{a+5}{a-5}$ е) $\frac{y^2-9}{y^2-6y+9}$ = $\frac{(y-3)(y+3)}{(y-3)(y-3)}$ = $\frac{y+3}{y-3}$ *Перевод:* a) $\frac{y^2-16}{3y+12}$ = $\frac{(y-4)(y+4)}{3(y+4)}$ = $\frac{y-4}{3}$ б) $\frac{5x-15y}{x^2-9y^2}$ = $\frac{5(x-3y)}{(x-3y)(x+3y)}$ = $\frac{5}{x+3y}$ в) $\frac{(c+2)^2}{7c^2+14c}$ = $\frac{(c+2)(c+2)}{7c(c+2)}$ = $\frac{c+2}{7c}$ г) $\frac{6cd-18c}{(d-3)^2}$ = $\frac{6c(d-3)}{(d-3)(d-3)}$ = $\frac{6c}{d-3}$ д) $\frac{a^2+10a+25}{a^2-25}$ = $\frac{(a+5)(a+5)}{(a-5)(a+5)}$ = $\frac{a+5}{a-5}$ е) $\frac{y^2-9}{y^2-6y+9}$ = $\frac{(y-3)(y+3)}{(y-3)(y-3)}$ = $\frac{y+3}{y-3}$