Пожалуйста помогите с СОРОМ (Даю 30 баллов)

Решите задачу:

$1) \ \frac{y^2+y}{y^2} = \frac{y(y+1)}{y^2} = \frac{y+1}{y} \\ \frac{a^2-2a}{a^2-4} = \frac{a(a-2)}{(a-2)(a+2)} = \frac{a}{a+2} \\ \\ 2) \ \frac{3}{a} + \frac{a-3}{a+5} = \frac{3(a+5)+a(a-3)}{a(a+5)} = \frac{3a+15+a^2-3a}{a(a+5)} = \frac{15+a^2}{a(a+5)} \\ \frac{2x^2}{x^2-4} - \frac{2x}{x+2} = \frac{2x^2-2x(x-2)}{(x-2)(x+2)} = \frac{2x^2-2x^2+4x}{(x-2)(x+2)} = \frac{4x}{(x-2)(x+2)} \\ \\$
$3) \ \frac{y^2-6y+9}{y^2-9} : \frac{10y-30}{y^2+3y} = \frac{(y-3)^2}{(y-3)(y+3)} : \frac{10(y-3)}{y(y+3)} = \frac{(y-3)^2}{(y-3)(y+3)} \cdot \frac{y(y+3)}{10(y-3)} = \\ = \frac{y}{10} \\ \\ 4) \ \frac{x}{y} = a \ ; \ \frac{y}{z} = \frac{1}{a} \ ; \ \frac{x}{z} = ? \\ x=ay \ ; \ z = ay \\ \frac{x}{z} = \frac{ay}{ay} = 1$