Вычислите (1;2; 4)........

Решите задачу:

$(42\frac{5}{12}-21 \frac{11}{18})-(25-4 \frac{1}{9})=$
$42 \frac{5}{12}-21 \frac{11}{18}= 42\frac{15}{36}-21 \frac{22}{36}=41 \frac{51}{36}-21\frac{22}{36}=20 \frac{29}{36}$
$25-4 \frac{1}{9}=24 \frac{9}{9}-4 \frac{1}{9}=20 \frac{8}{9}$
$20 \frac{29}{36}-20 \frac{8}{9}=20 \frac{29}{36}-20 \frac{32}{36}=-(20 \frac{32}{36}-20 \frac{29}{36})=- \frac{3}{36}= -\frac{1}{12}$
$(2 \frac{1}{2}:3 \frac{2}{3}):(7 \frac{1}{2}:7 \frac{1}{3})*(5 \frac{1}{4}* \frac{4}{21})$
$2 \frac{1}{2}:3 \frac{2}{3}= \frac{5}{2}: \frac{11}{3}= \frac{5}{2}* \frac{3}{11}= \frac{15}{22}$
$7 \frac{1}{2}:7 \frac{1}{3}= \frac{15}{2}: \frac{22}{3}= \frac{15}{2}* \frac{3}{22}= \frac{45}{44}$ $5 \frac{1}{4}* \frac{4}{21}= \frac{21}{4}* \frac{4}{21}=1$
$\frac{15}{22}: \frac{45}{44}= \frac{15}{22}* \frac{44}{45}= \frac{2}{3}$