Ответ:

а) \(4\frac{11}{18}\cdot\frac67-1\frac49=\frac{83}{18}\cdot\frac67-\frac{13}{9}=\frac{83}{21}-\frac{91}{63}=\frac{158}{63}=2\frac{32}{63}\).

б) \((1\frac12)^3-2\frac14\cdot1\frac18=\frac{27}{8}-\frac94\cdot\frac98=\frac{108}{32}-\frac{81}{32}=\frac{27}{32}\).

в) \((1\frac14)^2-\frac58=\frac{25}{16}-\frac{10}{16}=\frac{15}{16}\). Далее \(\frac{15}{16}\cdot10\frac23-7\frac13=\frac{15}{16}\cdot\frac{32}{3}-\frac{22}{3}=10-\frac{22}{3}=\frac83\).

г) В первых скобках \(1\frac49+2\frac56-2\frac34=\frac{52+102-99}{36}=\frac{55}{36}\). Во вторых \(2\frac12-\frac{11}{14}=\frac{12}{7}\). Поэтому \(\frac{55}{36}\cdot\frac{12}{7}=\frac{55}{21}=2\frac{13}{21}\).

Ответ: а) \(2\frac{32}{63}\); б) \(\frac{27}{32}\); в) \(\frac83\); г) \(2\frac{13}{21}\).