1.29. Сократите дроби: а) \(\frac{7a^2}{a^2(3a-9)(a+17)}\); б) \(\frac{3b+4}{(2b+1)(9b^2-16)}\); в) \(\frac{73c^2}{c^3(c+8)(13c-39)}\); г) \(\frac{2d-1}{(4d^2-1)(7d+5)}\).

1. \(\frac{7a^2}{a^2(3a-9)(a+17)}=\frac7{(3a-9)(a+17)}=\frac7{3(a-3)(a+17)}\), \(a\ne0,3,-17\).
2. \(9b^2-16=(3b-4)(3b+4)\), поэтому \(\frac{3b+4}{(2b+1)(3b-4)(3b+4)}=\frac1{(2b+1)(3b-4)}\), \(b\ne-\frac12,\pm\frac43\).
3. \(c^3=c^2\cdot c\), \(13c-39=13(c-3)\), поэтому \(\frac{73c^2}{c^3(c+8)(13c-39)}=\frac{73}{13c(c+8)(c-3)}\), \(c\ne0,-8,3\).
4. \(4d^2-1=(2d-1)(2d+1)\), поэтому \(\frac{2d-1}{(2d-1)(2d+1)(7d+5)}=\frac1{(2d+1)(7d+5)}\), \(d\ne\pm\frac12,-\frac57\).

Ответ: \(\frac7{3(a-3)(a+17)};\ \frac1{(2b+1)(3b-4)};\ \frac{73}{13c(c+8)(c-3)};\ \frac1{(2d+1)(7d+5)}\).