6) Сократи дроби: а) \(\frac{7,2\cdot2,8}{4,9\cdot6,4}\); б) \(\frac{12\cdot7-12\cdot3}{12\cdot7+12\cdot3}\); в) \(\frac{32abn^4}{96a^2b^3n^2}\).

а) \[\frac{7.2\cdot2.8}{4.9\cdot6.4}=\frac{72\cdot28}{49\cdot64}=\frac{9\cdot4}{7\cdot8}=\frac{9}{14}.\]

б) \[\frac{12\cdot7-12\cdot3}{12\cdot7+12\cdot3}=\frac{12(7-3)}{12(7+3)}=\frac4{10}=\frac25.\]

в) \[\frac{32abn^4}{96a^2b^3n^2}=\frac{1}{3ab^2}n^2=\frac{n^2}{3ab^2}.\]

Ответ: а) \(\frac9{14}\); б) \(\frac25\); в) \(\frac{n^2}{3ab^2}\).