б) (\frac{5y^2}{1-y^2}):(1-\frac{1}{1-y});

б)

$$(\frac{5y^2}{1-y^2}):(1-\frac{1}{1-y}) = (\frac{5y^2}{1-y^2}):(\frac{1-y-1}{1-y}) = (\frac{5y^2}{1-y^2}):(\frac{-y}{1-y}) = \frac{5y^2(1-y)}{(1-y^2)(-y)} = \frac{5y^2(1-y)}{(1-y)(1+y)(-y)} = \frac{5y}{-(1+y)} = -\frac{5y}{1+y}$$.

Ответ: $$\frac{-5y}{1+y}$$