Решите оба пожалуйста

№1
$1) \frac{7}{8}p *\frac{4}{9}k= \frac{7}{2}p* \frac{1}{9}k= \frac{7*p*k}{18}$
$2) \frac{17}{12}x* \frac{9}{51}y= \frac{1}{4}x* \frac{3}{3}y= \frac{1}{4}xy$
$3) \frac{21}{8}x*2y* \frac{16}{7}z= \frac{21}{8}x*y*\frac{8}{7}*z=3xyz$
№2
$1) \frac{5}{18}c + \frac{7}{24}c+ \frac{11}{30}c=( \frac{5}{3} + \frac{7}{4}+ \frac{11}{5})* \frac{1}{6} c= (\frac{100+105+132}{60})\frac{1}{6}c= \frac{337}{60} * \frac{1}{6}*c$ $= \frac{337}{360} c$
$2) \frac{8}{13}m+ \frac{5}{26}m+ \frac{7}{39} m=(8+ \frac{5}{2}+ \frac{7}{3})* \frac{1}{13}m=( \frac{48+15+14}{6}) * \frac{1}{13}m= \frac{77}{6}*$ $\frac{1}{13}m= \frac{77}{78}m$
$3) \frac{39}{16}y+ \frac{65}{20}y+ \frac{11}{8}y=( \frac{16}{2}+ 13+ \frac{11}{2})* \frac{1}{4}y= ( \frac{16+26+11}{2}) *\frac{1}{4}y= \frac{53}{8}y$