Ничего не получается. Прошу, помогите.

Решите задачу:

$1)\quad \frac{4c+3y}{c^2-y^2} - \frac{3}{c-y} = \frac{4c+3y}{(c-y)(c+y)} - \frac{3}{c-y} = \frac{4c+3y-3(c+y)}{(c-y)(c+y)} =\\\\= \frac{c}{(c-y)(c+y)} \\\\2)\quad \frac{c}{(c-y)(c+y)} + \frac{1}{y-c} = \frac{c}{(c-y)(c+y)} - \frac{1}{c-y} = \\\\=\frac{c-(c+y)}{(c-y)(c+y)} = \frac{-y}{(c-y)(c+y)} \\\\3)\quad \frac{-y}{(c-y)(c+y)y} - \frac{4}{5y+5c} = \frac{-y}{(c-y)(c+y)} - \frac{4}{5(c+y)} =\\\\= \frac{-5y-4(c-y)}{5(c-y)(c+y)} = \frac{-y-4c}{5(c-y)(c+y)} =- \frac{y+4c}{5(c^2-y^2)}$