Ответ:
- \(y^2-16=(y-4)(y+4)\).
\[\frac{6}{(y-4)(y+4)}-\frac{3(y+4)}{(y-4)(y+4)}=\frac{-3y-6}{y^2-16}.\]
Ограничения: \(y\ne\pm4\). - \(y^2-9=(y-3)(y+3)\).
\[\frac{y+4}{(y-3)(y+3)}-\frac{y-3}{(y-3)(y+3)}=\frac{7}{y^2-9}.\]
Ограничения: \(y\ne\pm3\). - \(d^2-25=(d-5)(d+5)\).
\[\frac{d}{(d-5)(d+5)}-\frac{d+5}{(d-5)(d+5)}=\frac{-5}{d^2-25}.\]
Ограничения: \(d\ne\pm5\).
