2/3.731, 32 24 M(15元-2층)-(5동+6号)+(10号-5층) N2a) 11/4-x=3元 8)+条-第二号-1/4 E

№1. Вычислите:

1. $$11\frac{1}{4}-x=3\frac{7}{10}$$
2. $$y+\frac{5}{7}-\frac{1}{8}=\frac{2}{3}-\frac{1}{14}$$

Решение:

№1

$$ (15\frac{1}{2}-2\frac{3}{8})-(5\frac{5}{6}+6\frac{3}{4})+(10\frac{2}{3}-5\frac{5}{8}) = $$

$$ = (\frac{31}{2}-\frac{19}{8})-(\frac{35}{6}+\frac{27}{4})+(\frac{32}{3}-\frac{45}{8}) = $$

$$ = (\frac{31 \cdot 4}{2 \cdot 4}-\frac{19}{8})-(\frac{35 \cdot 2}{6 \cdot 2}+ \frac{27 \cdot 3}{4 \cdot 3})+(\frac{32 \cdot 8}{3 \cdot 8}-\frac{45 \cdot 3}{8 \cdot 3}) = $$

$$ = (\frac{124}{8}-\frac{19}{8})-(\frac{70}{12}+ \frac{81}{12})+(\frac{256}{24}-\frac{135}{24}) = $$

$$ = \frac{105}{8} - \frac{151}{12} + \frac{121}{24} = \frac{105 \cdot 3}{8 \cdot 3} - \frac{151 \cdot 2}{12 \cdot 2} + \frac{121}{24} = $$

$$ = \frac{315}{24} - \frac{302}{24} + \frac{121}{24} = \frac{315-302+121}{24} = \frac{13+121}{24} = \frac{134}{24} = \frac{67}{12} = 5\frac{7}{12}$$

№2 a)

$$11\frac{1}{4}-x=3\frac{7}{10}$$

$$x = 11\frac{1}{4}-3\frac{7}{10}$$

$$x = \frac{45}{4} - \frac{37}{10}$$

$$x = \frac{45 \cdot 5}{4 \cdot 5} - \frac{37 \cdot 2}{10 \cdot 2}$$

$$x = \frac{225}{20} - \frac{74}{20}$$

$$x = \frac{151}{20} = 7\frac{11}{20}$$

№2 б)

$$y+\frac{5}{7}-\frac{1}{8}=\frac{2}{3}-\frac{1}{14}$$

$$y = \frac{2}{3}-\frac{1}{14} - \frac{5}{7}+\frac{1}{8}$$

$$y = \frac{2 \cdot 56}{3 \cdot 56} - \frac{1 \cdot 12}{14 \cdot 12} - \frac{5 \cdot 24}{7 \cdot 24} + \frac{1 \cdot 21}{8 \cdot 21}$$

$$y = \frac{112}{168} - \frac{12}{168} - \frac{120}{168} + \frac{21}{168}$$

$$y = \frac{112-12-120+21}{168} = \frac{100-120+21}{168} = \frac{-20+21}{168} = \frac{1}{168}$$

Ответ: №1 $$5\frac{7}{12}$$, №2 a) $$7\frac{11}{20}$$, б) $$\frac{1}{168}$$