$(\frac{16}{27}-\frac{5}{9})\ast \frac{3}{10}=\frac{1}{90}$
а) $\frac{16}{27}-\frac{5}{9}=\frac{16-5\ast 3}{27}=\frac{16-15}{27}=\frac{1}{27}$
б) $\frac{1}{27}\ast \frac{3}{10}=\frac{1\ast 3}{27\ast 10}=\frac{1\ast 1}{9\ast 10}=\frac{1}{90}$

$\frac{16}{27}-\frac{5}{9}\ast \frac{3}{10}=\frac{23}{54}$
а) $\frac{5}{9}\ast \frac{3}{10}=\frac{5\ast 3}{9\ast 10}=\frac{1\ast 1}{3\ast 2}=\frac{1}{6}$
б) $\frac{16}{27}-\frac{1}{6}=\frac{16\ast 2-1\ast 9}{54}=\frac{32-9}{54}=\frac{23}{54}$

$(6-2\frac{2}{7}\ast 1\frac{5}{44})\ast 2\frac{17}{19}=10$
а) $2\frac{2}{7}\ast 1\frac{5}{44}=\frac{16}{7}\ast \frac{49}{44}=\frac{16\ast 49}{7\ast 44}=\frac{4\ast 7}{1\ast 11}=\frac{28}{11}=2\frac{6}{11}$
б) $6-2\frac{6}{11}=5\frac{11}{11}-2\frac{6}{11}=3\frac{5}{11}$
в) $3\frac{5}{11}\ast 2\frac{17}{19}=\frac{38}{11}\ast \frac{55}{19}=\frac{38\ast 55}{11\ast 19}=\frac{2\ast 5}{1\ast 1}=10$