в) $$ \frac{d+3}{cd+d^2} - \frac{c-3}{cd+c^2} = \frac{d+3}{d(c+d)} - \frac{c-3}{c(c+d)} = \frac{c(d+3) - d(c-3)}{cd(c+d)} = \frac{cd+3c - cd + 3d}{cd(c+d)} = \frac{3c + 3d}{cd(c+d)} = \frac{3(c+d)}{cd(c+d)} = \frac{3}{cd} $$
Ответ: $$ \frac{3}{cd} $$