[m]\frac{8d–3}{3d(d–c)} - \frac{24c–9}{9c(d–c)} = \frac{8d–3}{3d(d–c)} - \frac{8c–3}{3c(d–c)} = \frac{c(8d–3)}{3cd(d–c)} - \frac{d(8c–3)}{3cd(d–c)} =[/m]
[m]=\frac{c(8d–3) - d(8c–3)}{3cd(d–c)} =\frac{8cd–3c - 8cd+3d)}{3cd(d–c)} =\frac{–3c+3d)}{3cd(d–c)} = \frac{3(d-c)}{3cd(d–c)} =\frac{1}{cd}[/m]
Ответ: 1/(cd)