Решите уравнение:
а) $х + 2\frac{2}{11} = 5$;
б) $26\frac{5}{8} + a = 30$;
в) $n - 6\frac{5}{6} = \frac{2}{9}$;
г) $11\frac{1}{4} - х = 3\frac{7}{10}$;
д) $3\frac{11}{24} - х = 1\frac{1}{6} + 1\frac{1}{9}$;
е) $y + \frac{5}{7} - \frac{1}{8} = \frac{2}{3} - \frac{1}{14}$.
$х + 2\frac{2}{11} = 5$
$х = 5 - 2\frac{2}{11} = 4\frac{11}{11} - 2\frac{2}{11} = 2\frac{9}{11}$
$26\frac{5}{8} + a = 30$
$a = 30 - 26\frac{5}{8} = 29\frac{8}{8} - 26\frac{5}{8} = 3\frac{3}{8}$
$n - 6\frac{5}{6} = \frac{2}{9}$
$n = \frac{2}{9} + 6\frac{5}{6} = \frac{4}{18} + 6\frac{15}{18} = 6\frac{19}{18} = 7\frac{1}{18}$
$11\frac{1}{4} - х = 3\frac{7}{10}$
$-х = 3\frac{7}{10} - 11\frac{1}{4}$
$х = 11\frac{1}{4} - 3\frac{7}{10} = 11\frac{5}{20} - 3\frac{14}{20} = 10\frac{25}{20} - 3\frac{14}{20} = 7\frac{11}{20}$
$3\frac{11}{24} - х = 1\frac{1}{6} + 1\frac{1}{9}$
$-х = 1\frac{1}{6} + 1\frac{1}{9} - 3\frac{11}{24}$
$х = 3\frac{11}{24} - 1\frac{1}{6} - 1\frac{1}{9} = 3\frac{33}{72} - 1\frac{12}{72} - 1\frac{8}{72} = 1\frac{13}{72}$
$y + \frac{5}{7} - \frac{1}{8} = \frac{2}{3} - \frac{1}{14}$
$y = \frac{2}{3} - \frac{1}{14} - \frac{5}{7} + \frac{1}{8} = \frac{112}{168} - \frac{12}{168} - \frac{120}{168} + \frac{21}{168} = \frac{1}{168}$
Пожаулйста, оцените решение
