Homer 939 (ж з и к) u Homer 940

Question 1

Question 2

Такие многоэтажные дроби - это деление
939. ж) $\frac{9}{16} : ( \frac{3}{4} - \frac{3}{8} )= \frac{9}{16}:( \frac{6}{8} - \frac{3}{8} )= \frac{9}{16}: \frac{3}{8} = \frac{9}{16}* \frac{8}{3} = \frac{3}{2}$
з) $( \frac{5}{6} - \frac{2}{3} ): \frac{1}{3} =( \frac{5}{6} - \frac{4}{6} ): \frac{1}{3} = \frac{1}{6} : \frac{1}{3} = \frac{1}{6} *3= \frac{3}{6} = \frac{1}{2}$
и) $( \frac{1}{2} + \frac{1}{6} ):( \frac{1}{2} - \frac{1}{6} )=( \frac{3}{6} + \frac{1}{6} ):( \frac{3}{6} - \frac{1}{6} )= \frac{4}{6} : \frac{2}{6} = \frac{4}{6} * \frac{6}{2} =2$
к) $( \frac{17}{14} - \frac{5}{7} ):( \frac{11}{21} + \frac{1}{7} )=( \frac{17}{14} - \frac{10}{14} ):( \frac{11}{21} + \frac{3}{21} )= \frac{7}{14}: \frac{14}{21} = \frac{1}{2}* \frac{3}{2} =3$

940. а) $[( \frac{3}{4} - \frac{1}{3} ): \frac{5}{7} ]:[( \frac{1}{4} + \frac{2}{3} )* \frac{6}{11} ]=( \frac{5}{12} : \frac{5}{7} ):( \frac{11}{12} * \frac{6}{11} )= \frac{7}{12} : \frac{1}{2} = \frac{7}{6}$
б) $[ \frac{3}{20}*( \frac{7}{12} - \frac{1}{2} )+ \frac{79}{80} ]:[\frac{13}{24}:( \frac{7}{12} + \frac{1}{2} )- \frac{1}{4} ] =[ \frac{3}{20}*\frac{1}{12}+ \frac{79}{80} ]:[\frac{13}{24}: \frac{13}{12}- \frac{1}{4} ]$
$=[ \frac{1}{80}+ \frac{79}{80} ]:[\frac{12}{24}- \frac{1}{4} ]=1:( \frac{1}{2} - \frac{1}{4} )=1: \frac{1}{4} =4$
в) $[(3+ \frac{7}{11} )* \frac{1}{4}- \frac{1}{22} ]:[(5- \frac{3}{11} ):13+ \frac{1}{2} ]=[ \frac{40}{11}* \frac{1}{4} - \frac{1}{22}]:[ \frac{52}{11*13}+ \frac{1}{2} ]$
$=( \frac{10}{11} - \frac{1}{22}):( \frac{4}{11}+ \frac{1}{2} )=( \frac{20}{22} - \frac{1}{22}):( \frac{8}{22}+ \frac{11}{22} )= \frac{19}{22} : \frac{19}{22} =1$