Вычислите: И) (0,4+8·(5−0,8·5/8)−5:2 1/2)/((1 7/8·8−(8,9−2,6:2/3))·34 2/5)·90; Й) 2−(3 1/3·1,9+19,5:4 1/2)/(62/75−0,16); К) ((6 3/5−3 3/14)·5 5/6)/((21−1,25):2,5); Л) 10/(1/2+3/4); М) (1/3−1/5)/(2/3−1/2); Н) 2−2/(3+1/2); О) (1−1/(3/4+1))/3; П) 1/(1+1/(1+1/(1+1/3))); Р) [(3 2/5+1 7/10)·1 3/17−(2 7/23−1 45/46)·69/80]·1 1/3.

И)

Числитель: \(0.4+8\cdot(5-0.8\cdot\frac{5}{8})-5:2.5=0.4+8\cdot4.5-2=34.4\).

Знаменатель: \((1\frac{7}{8}\cdot8-(8.9-2.6:\frac{2}{3}))\cdot34\frac{2}{5}=(15-(8.9-3.9))\cdot34.4=10\cdot34.4=344\).

\(\frac{34.4}{344}\cdot90=0.1\cdot90=9\).

Й)

\(3\frac{1}{3}\cdot1.9+19.5:4\frac{1}{2}=\frac{10}{3}\cdot\frac{19}{10}+\frac{39}{2}:\frac{9}{2}=\frac{19}{3}+\frac{13}{3}=\frac{32}{3}\).

\(\frac{62}{75}-0.16=\frac{62}{75}-\frac{4}{25}=\frac{2}{3}\).

\(2-\frac{32/3}{2/3}=2-16=-14\).

К)

\(6\frac{3}{5}-3\frac{3}{14}=\frac{33}{5}-\frac{45}{14}=\frac{237}{70}\).

\(\frac{237}{70}\cdot5\frac{5}{6}=\frac{237}{70}\cdot\frac{35}{6}=\frac{79}{4}\).

\((21-1.25):2.5=19.75:2.5=\frac{79}{10}\).

\(\frac{79/4}{79/10}=\frac{5}{2}\).

Л)

\(\frac{10}{\frac12+\frac34}=\frac{10}{\frac54}=8\).

М)

\(\frac{\frac13-\frac15}{\frac23-\frac12}=\frac{\frac{2}{15}}{\frac16}=\frac45\).

Н)

\(2-\frac{2}{3+\frac12}=2-\frac{2}{\frac72}=2-\frac47=\frac{10}{7}\).

О)

\(\frac{1-\frac{1}{\frac34+1}}{3}=\frac{1-\frac47}{3}=\frac{\frac37}{3}=\frac17\).

П)

\(1+\frac13=\frac43\), \(1+\frac{1}{4/3}=\frac74\), \(1+\frac{1}{7/4}=\frac{11}{7}\).

Следовательно, \(\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac13}}}=\frac{1}{11/7}=\frac{7}{11}\).

Р)

\(3\frac25+1\frac7{10}=\frac{51}{10}\), поэтому \(\frac{51}{10}\cdot1\frac3{17}=\frac{51}{10}\cdot\frac{20}{17}=6\).

\(2\frac7{23}-1\frac{45}{46}=\frac{106}{46}-\frac{91}{46}=\frac{15}{46}\).

\(\frac{15}{46}\cdot\frac{69}{80}=\frac{9}{32}\).

\(\left(6-\frac{9}{32}\right)\cdot1\frac13=\frac{183}{32}\cdot\frac43=\frac{61}{8}\).

Ответ: И) 9; Й) −14; К) \(\frac52\); Л) 8; М) \(\frac45\); Н) \(\frac{10}{7}\); О) \(\frac17\); П) \(\frac7{11}\); Р) \(\frac{61}{8}\).