ГДЗ Математика 5 класс Никольский, Потапов, Решетников

ГДЗ Математика 5 класс Никольский, Потапов, Решетников

авторы: Никольский, Потапов, Решетников.

издательство: Просвещение 2017 год

Математика 5 класс Никольский. Номер №1029

Вычислите:
а)

\frac{20 : 2 \frac{2}{15} + 25 \frac{5}{7} : 1 \frac{1}{35}}{21 \frac{7}{9} : 4 \frac{2}{3} - 1}

;
б)

\frac{6 \frac{3}{4} : 9 + 24 : \frac{6}{7} - \frac{1}{9} : \frac{4}{21}}{53 \frac{2}{3} - 22 \frac{14}{15} : 2 \frac{2}{3}}

;
в)

\frac{20 \frac{3}{10} - (2 \frac{1}{10} * 4 + \frac{19}{20} * 10)}{1 : 5}

;
г)

\frac{4 \frac{2}{5} - 2 \frac{3}{4} + 7 \frac{7}{15} - 8 \frac{7}{60}}{7 \frac{1}{4} - 5 \frac{3}{4}}

;
д)

\frac{(\frac{1}{6} + \frac{1}{10} + \frac{1}{15}) : (\frac{1}{6} + \frac{1}{30})}{(\frac{1}{6} + \frac{1}{4} - \frac{1}{5}) : (\frac{1}{4} - \frac{1}{6})}

;
е)

\frac{5 \frac{1}{5} : \frac{39}{40}}{2 \frac{4}{5} : \frac{7}{10} * \frac{3}{4}} + \frac{(2 - 1 \frac{3}{20}) * 48}{\frac{1}{4} * 20 + 1 : 10}

.

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Математика 5 класс Никольский. Номер №1029

Решение а

\frac{20 : 2 \frac{2}{15} + 25 \frac{5}{7} : 1 \frac{1}{35}}{21 \frac{7}{9} : 4 \frac{2}{3} - 1} = \frac{\frac{20}{1} * \frac{15}{32} + \frac{180}{7} * \frac{35}{36}}{\frac{196}{9} * \frac{3}{14} - 1} = \frac{\frac{4 * 5}{1} * \frac{15}{4 * 8} + \frac{36 * 5}{7} * \frac{7 * 5}{36}}{\frac{14 * 14}{3 * 3} * \frac{3}{14} - 1} = \frac{\frac{75}{8} + 25}{\frac{14}{3} - 1} = \frac{\frac{78}{8} + \frac{200}{8}}{\frac{14}{3} - \frac{3}{3}} = \frac{\frac{275}{8}}{\frac{11}{3}} = \frac{275}{8} * \frac{3}{11} = \frac{25 * 11}{8} * \frac{3}{11} = \frac{25}{8} * \frac{3}{1} = \frac{75}{8} = 9 \frac{3}{8}

Решение б

\frac{6 \frac{3}{4} : 9 + 24 : \frac{6}{7} - \frac{1}{9} : \frac{4}{21}}{53 \frac{2}{3} - 22 \frac{14}{15} : 2 \frac{2}{3}} = \frac{\frac{27}{4} * \frac{1}{9} + \frac{24}{1} * \frac{7}{6} - \frac{1}{9} * \frac{21}{4}}{53 \frac{2}{3} - \frac{344}{15} * \frac{3}{8}} = \frac{\frac{9 * 3}{4} * \frac{1}{9} + \frac{6 * 4}{1} * \frac{7}{6} - \frac{1}{3 * 3} * \frac{3 * 7}{4}}{53 \frac{2}{3} - \frac{43 * 8}{3 * 5} * \frac{3}{8}} = \frac{\frac{3}{4} + 28 - \frac{7}{12}}{53 \frac{2}{3} - \frac{43}{5}} = \frac{28 \frac{9}{12} - \frac{7}{12}}{53 \frac{2}{3} - 8 \frac{3}{5}} = \frac{28 \frac{2}{12}}{45 \frac{10 - 9}{15}} = \frac{28 \frac{1}{6}}{45 \frac{1}{15}} = \frac{169}{6} * \frac{15}{676} = \frac{169}{3 * 2} * \frac{3 * 5}{4 * 169} = \frac{5}{8}

Решение в

\frac{20 \frac{3}{10} - (2 \frac{1}{10} * 4 + \frac{19}{20} * 10)}{1 : 5} = \frac{20 \frac{3}{10} - (\frac{21}{10} * 4 + \frac{19}{20} * 10)}{1 * \frac{1}{5}} = \frac{20 \frac{3}{10} - (\frac{21 * 2 * 2}{2 * 5} + \frac{19 * 10}{2 * 10})}{\frac{1}{5}} = \frac{20 \frac{3}{10} - (\frac{42}{5} + \frac{19}{2})}{\frac{1}{5}} = \frac{20 \frac{3}{10} - \frac{84 + 95}{10}}{\frac{1}{5}} = \frac{20 \frac{3}{10} - \frac{179}{10}}{\frac{1}{5}} = \frac{19 \frac{13}{10} - 17 \frac{9}{10}}{\frac{1}{5}} = 2 \frac{4}{10} * 5 = 10 \frac{20}{10} = 12

Решение г

\frac{4 \frac{2}{5} - 2 \frac{3}{4} + 7 \frac{7}{15} - 8 \frac{7}{60}}{7 \frac{1}{4} - 5 \frac{3}{4}} = \frac{4 \frac{24}{60} - 2 \frac{45}{60} + 7 \frac{28}{60} - 8 \frac{7}{60}}{6 \frac{5}{4} - 5 \frac{3}{4}} = \frac{3 \frac{84}{60} - 2 \frac{45}{60} + 7 \frac{28}{60} - 8 \frac{7}{60}}{1 \frac{2}{4}} = \frac{1 \frac{39}{60} + 7 \frac{28}{60} - 8 \frac{7}{60}}{1 \frac{1}{2}} = \frac{8 \frac{67}{60} - 8 \frac{7}{60}}{\frac{3}{2}} = \frac{8 \frac{60}{60}}{\frac{3}{2}} = 9 * \frac{2}{3} = \frac{18}{3} = 6

Решение д

\frac{(\frac{1}{6} + \frac{1}{10} + \frac{1}{15}) : (\frac{1}{6} + \frac{1}{30})}{(\frac{1}{6} + \frac{1}{4} - \frac{1}{5}) : (\frac{1}{4} - \frac{1}{6})} = \frac{(\frac{5}{30} + \frac{3}{30} + \frac{2}{30}) : (\frac{5}{30} + \frac{1}{30})}{(\frac{10}{60} + \frac{15}{60} - \frac{12}{60}) : (\frac{6}{24} - \frac{4}{24})} = \frac{\frac{10}{30} : \frac{6}{30}}{\frac{13}{60} : \frac{2}{24}} = \frac{\frac{10}{30} * \frac{30}{6}}{\frac{13}{60} * \frac{24}{2}} = \frac{\frac{10}{6}}{\frac{13 * 12}{12 * 5}} = \frac{5}{3} * \frac{5}{13} = \frac{25}{39}

Решение е

\frac{5 \frac{1}{5} : \frac{39}{40}}{2 \frac{4}{5} : \frac{7}{10} * \frac{3}{4}} + \frac{(2 - 1 \frac{3}{20}) * 48}{\frac{1}{4} * 20 + 1 : 10} = \frac{\frac{26}{5} * \frac{40}{39}}{\frac{14}{5} * \frac{10}{7} * \frac{3}{4}} + \frac{(1 \frac{20}{20} - 1 \frac{3}{20}) * 48}{\frac{20}{4} + 1 * \frac{1}{10}} = \frac{\frac{13 * 2}{5} * \frac{8 * 5}{13 * 3}}{\frac{7 * 2}{5} * \frac{5 * 2}{7} * \frac{3}{2 * 2}} + \frac{\frac{17}{20} * 48}{5 + \frac{1}{10}} = \frac{\frac{2}{1} * \frac{8}{3}}{3} + \frac{\frac{17}{4 * 5} * \frac{4 * 12}{1}}{5 \frac{1}{10}} = \frac{\frac{16}{3}}{3} + \frac{\frac{17 * 12}{5}}{\frac{51}{10}} = \frac{16}{3} * \frac{1}{3} + \frac{204}{5} * \frac{10}{51} = \frac{16}{9} + \frac{51 * 4}{5} * \frac{5 * 2}{51} = 1 \frac{7}{9} + 8 = 9 \frac{7}{9}