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Решите задачу:

$\frac{y}{by-2b^2}-\frac{2}{y^2+y-2by-2b}\cdot \Big (1+\frac{3y+y^2}{3+y}\Big )=\\\\=\frac{y}{b(y-2b)}-\frac{2}{y(y+1)-2b(y+1)}\cdot \frac{3+y+3y+y^2}{3+y}=\\\\=\frac{y}{b(y-2b)}-\frac{2}{(y+1)(y-2b)}\cdot \frac{y^2+4y+3}{3+y}=\\\\=\frac{y}{b(y-2b)}-\frac{2}{(y+1)(y-2b)}\cdot \frac{(y+3)(y+1)}{y+3}=\frac{y}{b(y-2b)}-\frac{2}{y-2b}=\\\\=\frac{y-2b}{b(y-2b)}=\frac{1}{b}$