Выполним действия, заменяя десятичные дроби обыкновенными.
а) \(4\frac{3}{5}=\frac{23}{5}\), \(2\frac{2}{5}=\frac{12}{5}\).
\((-15.64\cdot4\frac{3}{5}+7.1)\cdot2\frac{2}{5}=(-\frac{391}{25}\cdot\frac{23}{5}+\frac{71}{10})\cdot\frac{12}{5}=(-\frac{8993}{125}+\frac{887.5}{125})\cdot\frac{12}{5}=-\frac{3222}{5}\cdot\frac{12}{5}=-1546.56\).
б) \((0.4-\frac{11}{15})\cdot1\frac{2}{7}-(\frac{7}{18}-0.5):1\frac{1}{6}=(-\frac{1}{3})\cdot\frac{9}{7}-(-\frac{1}{9}):\frac{7}{6}=-\frac{3}{7}+\frac{2}{21}=-\frac{1}{3}\).
в) \(0.8\cdot\frac{37.12-5.6\cdot0.14}{1.21+3.4\cdot0.35}=0.8\cdot\frac{36.336}{2.4}=12.112\).
г) \((2.17\cdot3.7-1.83\cdot2.3):19.1=(8.029-4.209):19.1=3.82:19.1=0.2\).
д) \((\frac{2}{3})^2+\frac{5}{9}=\frac{4}{9}+\frac{5}{9}=1\).
е) \(5\frac{4}{19}\cdot3\frac{4}{7}+1\frac{15}{19}:\frac{7}{25}-1\frac{2}{3}=\frac{99}{19}\cdot\frac{25}{7}+\frac{34}{19}\cdot\frac{25}{7}-\frac{5}{3}=\frac{3325}{133}-\frac{5}{3}=\frac{9290}{399}\).
ё) \(\frac{(3\frac{1}{3}-2\frac{2}{3})\cdot2.4}{(15.5+4.5):2\frac{1}{2}}=\frac{\frac{2}{3}\cdot\frac{12}{5}}{20:\frac{5}{2}}=\frac{8}{5}:8=\frac{1}{5}=0.2\).
ж) \(5\frac{3}{7}-(2\frac{1}{2}+1\frac{1}{3})\cdot\frac{1}{6}=\frac{38}{7}-\frac{23}{6}\cdot\frac{1}{6}=\frac{38}{7}-\frac{23}{36}=\frac{1205}{252}\).
з) \((2\frac{7}{15}+1\frac{7}{12})\cdot1\frac{1}{9}-\frac{7}{8}=(\frac{37}{15}+\frac{19}{12})\cdot\frac{10}{9}-\frac{7}{8}=\frac{263}{54}-\frac{7}{8}=\frac{925}{216}\).
и) \((6\frac{1}{7}-5\frac{3}{4}):\frac{11}{14}+(3\frac{3}{4}-1\frac{5}{6}):\frac{1}{6}=\frac{11}{28}\cdot\frac{14}{11}+\frac{23}{12}\cdot6=\frac{1}{2}+\frac{23}{2}=12\).
Ответ: а) −1546.56; б) −1/3; в) 12.112; г) 0.2; д) 1; е) 9290/399; ё) 0.2; ж) 1205/252; з) 925/216; и) 12.