Помогите решить пример (5целых две третьи - 1 целая одна седьмая+3 целых десять двадцать...

Решите задачу:

$(5\frac{2}{3}-1 \frac{1}{7}+3 \frac{10}{21})* \frac{9}{16}+ \frac{6}{7}:(4- 1\frac{8}{11}+ \frac{1}{1}- \frac{17}{22})+3 \frac{1}{14}=\\ \\1)5\frac{2}{3}-1 \frac{1}{7}+3 \frac{10}{21}= \frac{5*3+2}{3}- \frac{7*1+1}{7}+ \frac{3*21+10}{21}= \frac{17*7}{3*7}- \frac{8*3}{7*3}+ \frac{73*1}{21*1}=\\\\= \frac{119-24+73}{21}= \frac{168:21}{21:21}= \frac{8}{1}=8;\\\\ 2)8* \frac{9}{16}= \frac{(8:8)*9}{16:8}= \frac{9}{2}=4 \frac{1}{2};\\\\$
$3)4- 1\frac{8}{11}+ \frac{1}{1}- \frac{17}{22}=4- \frac{19}{11}+1- \frac{17}{22}=5-( \frac{19*2}{11*2}+ \frac{17}{22})=5- \frac{38+17}{22}=\\\\=5- \frac{55}{22}=5-2 \frac{11}{22}=4 \frac{22}{22}-2 \frac{11}{22}=2 \frac{11}{22}= 2\frac{1}{2};\\\\ 4) \frac{6}{7}:2 \frac{1}{2}= \frac{6}{7}: \frac{5}{2}=\frac{6}{7}*\frac{2}{5}= \frac{12}{35};\\\\ 5) 4 \frac{1}{2}+ \frac{12}{35}= \frac{9*35}{2*35}+ \frac{12*2}{35*2}= \frac{315+24}{70}= \frac{339}{70}=4 \frac{59}{70};\\\\$
$6)4 \frac{59*1}{70*1}+3 \frac{1*5}{14*5}=4 \frac{59}{70}+3 \frac{5}{70}=7 \frac{64}{70}=7 \frac{32}{35}.$