Представьте дробь в виде неправильной дроби:
6\(\frac{2}{5}\) = \(\frac{6 \cdot 5 + 2}{5}\) = \(\frac{30 + 2}{5}\) = \(\frac{32}{5}\)
8\(\frac{1}{13}\) = \(\frac{8 \cdot 13 + 1}{13}\) = \(\frac{104 + 1}{13}\) = \(\frac{105}{13}\)
14\(\frac{11}{12}\) = \(\frac{14 \cdot 12 + 11}{12}\) = \(\frac{168 + 11}{12}\) = \(\frac{179}{12}\)
Переведите в смешанную дробь:
\(\frac{42}{7}\) = 6
\(\frac{60}{9}\) = 6\(\frac{6}{9}\) = 6\(\frac{2}{3}\)
\(\frac{91}{12}\) = 7\(\frac{7}{12}\)
Вычислите:
\(\frac{4}{7}\) + \(\frac{2}{5}\) = \(\frac{4 \cdot 5 + 2 \cdot 7}{35}\) = \(\frac{20 + 14}{35}\) = \(\frac{34}{35}\)
6\(\frac{3}{5}\) - 4 = 2\(\frac{3}{5}\)
Решите уравнения:
\(\frac{3}{5}\) + x = 2\(\frac{1}{4}\)
x = 2\(\frac{1}{4}\) - \(\frac{3}{5}\)
x = \(\frac{9}{4}\) - \(\frac{3}{5}\)
x = \(\frac{9 \cdot 5 - 3 \cdot 4}{20}\)
x = \(\frac{45 - 12}{20}\)
x = \(\frac{33}{20}\) = 1\(\frac{13}{20}\)
5\(\frac{1}{6}\) - x = 2\(\frac{1}{2}\)
x = 5\(\frac{1}{6}\) - 2\(\frac{1}{2}\)
x = \(\frac{31}{6}\) - \(\frac{5}{2}\)
x = \(\frac{31 - 5 \cdot 3}{6}\)
x = \(\frac{31 - 15}{6}\)
x = \(\frac{16}{6}\) = \(\frac{8}{3}\) = 2\(\frac{2}{3}\)
\(\frac{7}{15}\) * x = \(\frac{21}{25}\)
x = \(\frac{21}{25}\) : \(\frac{7}{15}\)
x = \(\frac{21}{25}\) * \(\frac{15}{7}\)
x = \(\frac{3}{5}\) * \(\frac{3}{1}\)
x = \(\frac{9}{5}\) = 1\(\frac{4}{5}\)
В первый день Артём прочитал \(\frac{9}{45}\) часть книги, а во второй день на \(\frac{4}{6}\) страниц больше. Какую часть книги Артём прочитал за 2 дня?
Решение:
Пусть вся книга - 1
\(\frac{9}{45}\) + \(\frac{4}{6}\) = \(\frac{9 \cdot 6 + 4 \cdot 45}{45 \cdot 6}\) = \(\frac{54 + 180}{270}\) = \(\frac{234}{270}\) = \(\frac{117}{135}\) = \(\frac{39}{45}\) - прочитал во второй день.
\(\frac{9}{45}\) + \(\frac{39}{45}\) = \(\frac{48}{45}\) = \(\frac{16}{15}\) = 1\(\frac{1}{15}\) - прочитал за два дня.
Ответ: 1\(\frac{1}{15}\)