2. Представьте в виде дроби: а) \(\frac{14a+8}{5a}-\frac{3-a^2}{a^2}\); б) \(\frac{5}{2y+x}-\frac{5}{2y-x}\); в) \(\frac{3m+7}{4m-m^2}+\frac{6}{4-m}\); г) \(\frac{2c+d}{8c-3d}-\frac{c-9d}{3d-8c}\).

1. а) Общий знаменатель — \(5a^2\): \[\frac{a(14a+8)-5(3-a^2)}{5a^2}=\frac{19a^2+8a-15}{5a^2}.\]
2. б) \[\frac{5(2y-x)-5(2y+x)}{(2y+x)(2y-x)}=\frac{-10x}{4y^2-x^2}.\]
3. в) Так как \(4m-m^2=m(4-m)\), \[\frac{3m+7}{m(4-m)}+\frac{6m}{m(4-m)}=\frac{9m+7}{m(4-m)}.\]
4. г) Поскольку \(3d-8c=-(8c-3d)\), \[\frac{2c+d}{8c-3d}+\frac{c-9d}{8c-3d}=\frac{3c-8d}{8c-3d}.\]

Ответ: а) \(\frac{19a^2+8a-15}{5a^2}\); б) \(-\frac{10x}{4y^2-x^2}\); в) \(\frac{9m+7}{m(4-m)}\); г) \(\frac{3c-8d}{8c-3d}\).