Учебник математики для 5-го класса под редакцией Петерсон – это современный и увлекательный подход к обучению математике. Он ориентирован на развитие логического мышления и творческого подхода к решению задач, что делает его особенно привлекательным для учеников.
ГДЗ по Математике 5 Класс Часть 2 Номер 603 Петерсон — Подробные Ответы
\(
\text{Реши уравнения и расшифруй название известной книги. Кто её написал?}
\)
А.
\(
2 \frac{1}{2} x — 1 \frac{5}{8} = 2 \frac{3}{4}
\)
Ю.
\(
(2 \frac{1}{10} — x) : 8 + 1 \frac{2}{15} = 1 \frac{1}{3}
\)
Н.
\(
\frac{1}{4} + \frac{x}{3 \frac{5}{9}} = \frac{7}{16}
\)
Р.
\(
4 \frac{1}{6} : (\frac{1}{4} x + 1 \frac{4}{15}) — 1 \frac{5}{6} = \frac{2}{3}
\)
Ь.
\(
4 \frac{2}{5} : x — 2 \frac{3}{5} = \frac{7}{10}
\)
Г.
\(
1 \frac{1}{2} x + \frac{1}{2} x = 1 \frac{2}{3}
\)
Л.
\(
3 \frac{1}{3} : (2 \frac{1}{8} — x) = 2 \frac{2}{9}
\)
П.
\(
3 \frac{4}{5} x — 1 \frac{7}{10} x = 3 \frac{1}{2}
\)
Э.
\(
(\frac{5}{12} + \frac{1}{2} x) : 2 \frac{1}{2} = \frac{11}{12}
\)
И.
\(
\frac{2}{3} x + \frac{7}{12} + \frac{1}{4} x + \frac{5}{6} = 5 \frac{1}{12}
\)
Ф.
\(
(6 \frac{3}{14} — x) \cdot 2 \frac{1}{3} = 9 \frac{5}{6}
\)
Т.
\(
\frac{3}{5} + \frac{4}{5} x + \frac{7}{10} x + 1 \frac{1}{2} = 5 \frac{3}{5}
\)
А
\(
2 \frac{1}{2} x — 1 \frac{5}{8} = 2 \frac{3}{4}
\)
\(
\frac{5}{2} x — \frac{13}{8} = \frac{11}{4} \quad | \cdot 8
\)
\(
5x \cdot 4 — 13 = 11 \cdot 2
\)
\(
20x — 13 = 22
\)
\(
20x = 22 + 13
\)
\(
20x = 35
\)
\(
x = \frac{35}{20}
\)
\(
x = \frac{7}{4}
\)
\(
x = 1 \frac{3}{4}
\)
Ю
\(
\left( 2 \frac{1}{10} — x \right) : 8 + 1 \frac{2}{15} = 1 \frac{1}{3}
\)
\(
\left( 2 \frac{1}{10} — x \right) : 8 = 1 \frac{1}{3} — 1 \frac{2}{15}
\)
\(
\left( 2 \frac{1}{10} — x \right) : 8 = 1 \frac{5}{15} — 1 \frac{2}{15}
\)
\(
\left( 2 \frac{1}{10} — x \right) : 8 = \frac{3}{15}
\)
\(
\left( 2 \frac{1}{10} — x \right) : 8 = \frac{1}{5}
\)
\(
2 \frac{1}{10} — x = \frac{1}{5} \cdot 8
\)
\(
2 \frac{1}{10} — x = \frac{8}{5}
\)
\(
2 \frac{1}{10} — x = 1 \frac{3}{5}
\)
\(
x = 2 \frac{1}{10} — 1 \frac{3}{5}
\)
\(
x = 1 \frac{11}{10} — 1 \frac{6}{10}
\)
\(
x = \frac{5}{10}
\)
\(
x = \frac{1}{2}
\)
Н
\(
\frac{1}{4} + x : 3 \frac{5}{9} = \frac{7}{16}
\)
\(
x : 3 \frac{5}{9} = \frac{7}{16} — \frac{1}{4}
\)
\(
x : 3 \frac{5}{9} = \frac{7}{16} — \frac{4}{16}
\)
\(
x : \frac{32}{9} = \frac{3}{16}
\)
\(
x = \frac{3}{16} \cdot \frac{32}{9}
\)
\(
x = \frac{3 \cdot 32}{16 \cdot 9}
\)
\(
x = \frac{2}{3}
\)
Р
\(
4 \frac{1}{6} : \left(\frac{1}{4} x + 1 \frac{4}{15}\right) — 1 \frac{5}{6} = \frac{2}{3}
\)
\(
4 \frac{1}{6} : \left(\frac{1}{4} x + 1 \frac{4}{15}\right) = \frac{2}{3} + 1 \frac{5}{6}
\)
\(
4 \frac{1}{6} : \left(\frac{1}{4} x + 1 \frac{4}{15}\right) = \frac{4}{6} + 1 \frac{5}{6}
\)
\(
4 \frac{1}{6} : \left(\frac{1}{4} x + 1 \frac{4}{15}\right) = 1 \frac{9}{6}
\)
\(
4 \frac{1}{6} : \left(\frac{1}{4} x + 1 \frac{4}{15}\right) = 1 \frac{3}{2}
\)
\(
\frac{1}{4} x + 1 \frac{4}{15} = 4 \frac{1}{6} : 1 \frac{3}{2}
\)
\(
\frac{1}{4} x + 1 \frac{4}{15} = \frac{25}{6} : \frac{5}{2}
\)
\(
\frac{1}{4} x + 1 \frac{4}{15} = \frac{25 \cdot 2}{6 \cdot 5}
\)
\(
\frac{1}{4} x + 1 \frac{4}{15} = \frac{5}{3}
\)
\(
\frac{1}{4} x = \frac{5}{3} — 1 \frac{4}{15}
\)
\(
\frac{1}{4} x = 1 \frac{2}{3} — 1 \frac{4}{15}
\)
\(
\frac{1}{4} x = 1 \frac{10}{15} — 1 \frac{4}{15}
\)
\(
\frac{1}{4} x = \frac{6}{15}
\)
\(
\frac{1}{4} x = \frac{2}{5}
\)
\(
x = \frac{2}{5} : \frac{1}{4}
\)
\(
x = \frac{2 \cdot 4}{5}
\)
\(
x = \frac{8}{5}
\)
\(
x = 1 \frac{3}{5}
\)
Б
\(
4 \frac{2}{5} : x — 2 \frac{3}{5} = \frac{7}{10}
\)
\(
4 \frac{2}{5} : x = \frac{7}{10} + 2 \frac{3}{5}
\)
\(
4 \frac{2}{5} : x = \frac{7}{10} + \frac{6}{10}
\)
\(
4 \frac{2}{5} : x = 2 \frac{13}{10}
\)
\(
x = 4 \frac{2}{5} : 2 \frac{13}{10}
\)
\(
x = \frac{22}{5} : \frac{33}{10}
\)
\(
x = \frac{22 \cdot 10}{5 \cdot 33}
\)
\(
x = \frac{2 \cdot 2}{1 \cdot 3}
\)
\(
x = \frac{4}{3}
\)
\(
x = 1 \frac{1}{3}
\)
Г
\(
1 \frac{1}{2} x + \frac{1}{2} x = 1 \frac{2}{3}
\)
\(
2x = \frac{5}{3}
\)
\(
2x \cdot 3 = 5
\)
\(
6x = 5
\)
\(
x = \frac{5}{6}
\)
Л
\(
3 \frac{1}{8} : (2 \frac{1}{8} — x) = 2 \frac{2}{9}
\)
\(
2 \frac{1}{8} — x = 3 \frac{1}{3} : 2 \frac{2}{9}
\)
\(
2 \frac{1}{8} — x = \frac{10}{3} : \frac{20}{9}
\)
\(
2 \frac{1}{8} — x = \frac{10 \cdot 9}{3 \cdot 20}
\)
\(
2 \frac{1}{8} — x = \frac{3}{2}
\)
\(
x = 2 \frac{1}{8} — \frac{3}{2}
\)
\(
x = 1 \frac{9}{8} — 1 \frac{4}{8}
\)
\(
x = \frac{5}{8}
\)
П
\(
3 \frac{4}{5} x — 1 \frac{7}{10} x = 3 \frac{1}{2}
\)
\(
3 \frac{8}{10} x — 1 \frac{7}{10} x = 3 \frac{1}{2}
\)
\(
2 \frac{1}{10} x = 3 \frac{1}{2}
\)
\(
x = 3 \frac{1}{2} : 2 \frac{1}{10}
\)
\(
x = \frac{7}{2} : \frac{21}{10}
\)
\(
x = \frac{7 \cdot 10}{2 \cdot 21}
\)
\(
x = \frac{5}{3}
\)
\(
x = 1 \frac{2}{3}
\)
Э
\(
\left(\frac{5}{12} + \frac{1}{2} x \right) : 2 \frac{1}{2} = \frac{11}{12}
\)
\(
\frac{5}{12} + \frac{1}{2} x = \frac{11}{12} \cdot 2 \frac{1}{2}
\)
\(
\frac{5}{12} + \frac{1}{2} x = \frac{11}{12} \cdot \frac{5}{2}
\)
\(
\frac{5}{12} + \frac{1}{2} x = \frac{55}{24}
\)
\(
\frac{1}{2} x = \frac{55}{24} — \frac{5}{12}
\)
\(
\frac{1}{2} x = \frac{55}{24} — \frac{10}{24}
\)
\(
\frac{1}{2} x = \frac{45}{24}
\)
\(
x = \frac{45}{24} : \frac{1}{2}
\)
\(
x = \frac{45}{24} \cdot 2
\)
\(
x = \frac{45}{12}
\)
\(
x = \frac{15}{4}
\)
\(
x = 3 \frac{3}{4}
\)
И
\(
\frac{2}{3} x + \frac{7}{12} + \frac{1}{4} x + \frac{5}{6} = 5 \frac{1}{12}
\)
\(
\left(\frac{2}{3} x + \frac{1}{4} x \right) + \left( \frac{7}{12} + \frac{5}{6} \right) = 5 \frac{1}{12}
\)
\(
\frac{8}{12} x + \frac{3}{12} x + \frac{7}{12} + \frac{10}{12} = 5 \frac{1}{12}
\)
\(
\frac{11}{12} x + \frac{17}{12} = 5 \frac{1}{12}
\)
\(
\frac{11}{12} x + 1 \frac{5}{12} = 5 \frac{1}{12}
\)
\(
\frac{11}{12} x = 5 \frac{1}{12} — 1 \frac{5}{12}
\)
\(
\frac{11}{12} x = 4 \frac{13}{12} — 1 \frac{5}{12}
\)
\(
\frac{11}{12} x = 3 \frac{8}{12}
\)
\(
\frac{11}{12} x = 3 \frac{2}{3}
\)
\(
x = 3 \frac{2}{3} : \frac{11}{12}
\)
\(
x = \frac{11 \cdot 12}{3 \cdot 11}
\)
\(
x = 4
\)
Ф
\(
\left(6 \frac{3}{14} — x \right) \cdot 2 \frac{1}{3} = 9 \frac{5}{6}
\)
\(
6 \frac{3}{14} — x = 9 \frac{5}{6} : 2 \frac{1}{3}
\)
\(
6 \frac{3}{14} — x = \frac{59}{7} : \frac{7}{3}
\)
\(
6 \frac{3}{14} — x = \frac{59 \cdot 3}{6 \cdot 7}
\)
\(
6 \frac{3}{14} — x = \frac{59}{2 \cdot 7}
\)
\(
6 \frac{3}{14} — x = \frac{59}{14}
\)
\(
6 \frac{3}{14} — x = 4 \frac{3}{14}
\)
\(
x = 6 \frac{3}{14} — 4 \frac{3}{14}
\)
\(
x = 2
\)
Т
\(
\frac{3}{5} + \frac{4}{5} x + \frac{7}{10} x + 1 \frac{1}{2} = 5 \frac{3}{5}
\)
\(
\left(\frac{4}{5} x + \frac{7}{10} x \right) + \left(\frac{3}{5} + 1 \frac{1}{2}\right) = 5 \frac{3}{5}
\)
\(
\left(\frac{8}{10} x + \frac{7}{10} x \right) + \left(\frac{6}{10} + \frac{15}{10}\right) = 5 \frac{3}{5}
\)
\(
\frac{15}{10} x + 1 \frac{11}{10} = 5 \frac{3}{5}
\)
\(
\frac{3}{2} x + 2 \frac{1}{10} = 5 \frac{3}{5}
\)
\(
\frac{3}{2} x = 5 \frac{3}{5} — 2 \frac{1}{10}
\)
\(
\frac{3}{2} x = 5 \frac{6}{10} — 2 \frac{1}{10}
\)
\(
\frac{3}{2} x = 3 \frac{5}{10}
\)
\(
\frac{3}{2} x = 3 \frac{1}{2}
\)
\(
\frac{3}{2} x = \frac{7}{2}
\)
\(
x = \frac{7}{2} : \frac{3}{2}
\)
\(
x = \frac{7 \cdot 2}{2 \cdot 3}
\)
\(
x = \frac{7}{3}
\)
\(
x = 2 \frac{1}{3}
\)
| \(\frac{5}{6}\) | 1 \(\frac{3}{4}\) | 1 \(\frac{3}{5}\) | \(\frac{5}{6}\) | 1 \(\frac{3}{4}\) | \(\frac{2}{3}\) | 2 \(\frac{1}{3}\) | \(\frac{1}{2}\) | 1 \(\frac{3}{4}\) | \(\frac{3}{4}\) | И |
| Г | А | Р | Г | А | Н | Т | Ю | А | И |
| 1 \(\frac{2}{3}\) | 1 \(\frac{3}{4}\) | \(\frac{2}{3}\) | 2 \(\frac{1}{3}\) | 1 \(\frac{3}{4}\) | \(\frac{5}{6}\) | 1 \(\frac{3}{5}\) | \(\frac{1}{2}\) | 3 \(\frac{3}{4}\) | \(\frac{5}{8}\) | 1 \(\frac{1}{3}\) |
| П | А | Н | Т | А | Г | Р | Ю | Э | Л | Ь |
Ответ:
«Гаргантюа и Пантагрюэль» — написал Франсуа Рабле.
