1) (4 1/6 - 2 7/8) * 24 / 1,6 + 1,5 = II B 2) 22,2 : 5 2/7 - 2 3/5 = 3) (7 1/4 - 6 7/18) * 7,2 + 2,8 = 4) 14/7 = x/45

Решение:

1) \[\frac{(4\frac{1}{6}-2\frac{7}{8})\cdot24}{1,6+1,5} = \frac{(\frac{25}{6}-\frac{23}{8})\cdot24}{3,1} = \frac{(\frac{100}{24}-\frac{69}{24})\cdot24}{3,1} = \frac{\frac{31}{24}\cdot24}{3,1} = \frac{31}{3,1} = 10\]

2) \[22,2 : 5\frac{2}{7} - 2\frac{3}{5} = 22,2 : \frac{37}{7} - \frac{13}{5} = \frac{222}{10} : \frac{37}{7} - \frac{13}{5} = \frac{111}{5} : \frac{37}{7} - \frac{13}{5} = \frac{111}{5} \cdot \frac{7}{37} - \frac{13}{5} = \frac{3\cdot7}{5} - \frac{13}{5} = \frac{21}{5} - \frac{13}{5} = \frac{8}{5} = 1\frac{3}{5}\]

3) \[(7\frac{1}{4}-6\frac{7}{18})\cdot7,2+2,8 = (7\frac{9}{36}-6\frac{14}{36})\cdot7,2+2,8 = (6\frac{45}{36}-6\frac{14}{36})\cdot7,2+2,8 = \frac{31}{36}\cdot7,2+2,8 = \frac{31}{36}\cdot\frac{72}{10}+2,8 = \frac{31}{1}\cdot\frac{2}{10}+2,8 = \frac{62}{10}+2,8 = 6,2+2,8 = 9\]

4) \[\frac{14}{7}=\frac{x}{45}\]

Воспользуемся основным свойством пропорции:

\[x = \frac{14\cdot45}{7} = 2\cdot45 = 90\]

Ответ: 1) 10; 2) 1 3/5; 3) 9; 4) 90.