Помогите решить(под буквой а)

$\frac{1}{4}*4 \frac{3}{4}* \frac{16}{57}+(4 \frac{3}{4}+1 \frac{2}{3})* \frac{16}{21}+ \frac{2}{27}*4 \frac{1}{2}=5 \frac{1}{3}$
1) $\frac{1}{4}*4 \frac{3}{4}* \frac{16}{57}= \frac{1*19*16}{4*4*57}= \frac{1}{3}$
2) $4 \frac{3}{4}+1 \frac{2}{3}=4 \frac{9}{12}+1 \frac{8}{12}=5 \frac{17}{12}=6 \frac{5}{12}$
3) $6 \frac{5}{12}* \frac{16}{21}= \frac{77*16}{12*21}= \frac{44}{9}=4 \frac{8}{9}$
4) $\frac{2}{27}*4 \frac{1}{2}= \frac{2*9}{27*2}= \frac{1}{3}$
5) $\frac{1}{3}+4 \frac{8}{9}+ \frac{1}{3}= \frac{3}{9}+4 \frac{8}{9}+ \frac{3}{9}=4 \frac{12}{9} =5 \frac{3}{9}= 5\frac{1}{3}$