ГДЗ по Математике 6 Класс Вариант 4 Номер 84 Мерзляк, Полонский — Подробные Ответы

Решите уравнение:

1)
\[
\frac{1}{3}x + \frac{1}{5}x + \frac{1}{6}x = \frac{21}{40}
\]

2)
\[
5 \frac{1}{4}x — 2 \frac{2}{3} = 1 \frac{5}{12}
\]

3)
\[
8 \frac{4}{15} — 5 \frac{2}{5}x = 4 \frac{2}{3}
\]

4)
\[
\frac{11}{24}x + 7 \frac{1}{3} = 9 \frac{5}{8}
\]

1)
\[
\left(\frac{1}{3} + \frac{1}{5} + \frac{1}{6}\right)x = \frac{21}{40}
\]

\[
\left(\frac{10}{30} + \frac{6}{30} + \frac{5}{30}\right)x = \frac{21}{40}
\]

\[
\frac{21}{30}x = \frac{21}{40}
\]

\[
x = \frac{21}{40} \cdot \frac{30}{21} = \frac{30}{40} = \frac{3}{4}
\]

2)
\[
5{,}\frac{1}{4}x — 2{,}\frac{2}{3} = 1{,}\frac{5}{12}
\]

\[
5{,}\frac{1}{4}x = 1{,}\frac{5}{12} + 2{,}\frac{2}{3}
\]

\[
5{,}\frac{1}{4}x = 1{,}\frac{5}{12} + 2{,}\frac{8}{12} = 3{,}\frac{13}{12}
\]

\[
x = \frac{3{,}\frac{13}{12}}{5{,}\frac{1}{4}}
\]

\[
x = \frac{\frac{49}{12}}{\frac{21}{4}} = \frac{49}{12} \cdot \frac{4}{21} = \frac{196}{252} = \frac{7}{9}
\]

3)
\[
8{,}\frac{4}{15} — 5{,}\frac{2}{5}x = 4{,}\frac{2}{3}
\]

\[
5{,}\frac{2}{5}x = 8{,}\frac{4}{15} — 4{,}\frac{2}{3}
\]

\[
8{,}\frac{4}{15} = \frac{124}{15},\quad 4{,}\frac{2}{3} = \frac{14}{3} = \frac{70}{15}
\]

\[
5{,}\frac{2}{5}x = \frac{124}{15} — \frac{70}{15} = \frac{54}{15}
\]

\[
x = \frac{54}{15} \div \frac{27}{5} = \frac{54}{15} \cdot \frac{5}{27} = \frac{270}{405} = \frac{2}{3}
\]

4)
\[
\frac{11}{24}x + 7{,}\frac{1}{3} = 9{,}\frac{5}{8}
\]

\[
\frac{11}{24}x = 9{,}\frac{5}{8} — 7{,}\frac{1}{3}
\]

\[
9{,}\frac{5}{8} = \frac{77}{8},\quad 7{,}\frac{1}{3} = \frac{22}{3} = \frac{66}{9}
\]

\[
\frac{77}{8} — \frac{22}{3} = \frac{231 — 176}{24} = \frac{55}{24}
\]

\[
\frac{11}{24}x = \frac{55}{24}
\]

\[
x = \frac{55}{24} \div \frac{11}{24} = 5
\]

1)
\[
\left( \frac{1}{3} + \frac{1}{5} + \frac{1}{6} \right)x = \frac{21}{40}
\]

Сначала складываем коэффициенты при \(x\), приводим к общему знаменателю:
\[
\frac{1}{3} = \frac{10}{30},\quad \frac{1}{5} = \frac{6}{30},\quad \frac{1}{6} = \frac{5}{30}
\]

\[
\left( \frac{10}{30} + \frac{6}{30} + \frac{5}{30} \right)x = \frac{21}{40}
\]

\[
\frac{21}{30}x = \frac{21}{40}
\]

Делим обе части на \(\frac{21}{30}\):
\[
x = \frac{21}{40} \div \frac{21}{30} = \frac{21}{40} \cdot \frac{30}{21}
\]

Сокращаем \(21\):
\[
x = \frac{30}{40}
\]

Сокращаем на 10:
\[
x = \frac{3}{4}
\]

2)
\[
5{,}\frac{1}{4}x — 2{,}\frac{2}{3} = 1{,}\frac{5}{12}
\]

Переносим \(-2{,}\frac{2}{3}\) вправо:
\[
5{,}\frac{1}{4}x = 1{,}\frac{5}{12} + 2{,}\frac{2}{3}
\]

Приводим смешанные числа к неправильным дробям:
\[
1{,}\frac{5}{12} = 1 + \frac{5}{12} = \frac{17}{12}
\]

\[
2{,}\frac{2}{3} = 2 + \frac{2}{3} = \frac{8}{3}
\]

\[
\frac{8}{3} = \frac{32}{12}
\]

Складываем:
\[
\frac{17}{12} + \frac{32}{12} = \frac{49}{12}
\]

\[
5{,}\frac{1}{4} = 5 + \frac{1}{4} = \frac{21}{4}
\]

Делим обе части на \(\frac{21}{4}\):
\[
x = \frac{49}{12} \div \frac{21}{4} = \frac{49}{12} \cdot \frac{4}{21}
\]

\[
x = \frac{196}{252}
\]

Сокращаем на 28:
\[
x = \frac{7}{9}
\]

3)
\[
8{,}\frac{4}{15} — 5{,}\frac{2}{5}x = 4{,}\frac{2}{3}
\]

Переносим \(5{,}\frac{2}{5}x\) вправо, а \(4{,}\frac{2}{3}\) влево:
\[
5{,}\frac{2}{5}x = 8{,}\frac{4}{15} — 4{,}\frac{2}{3}
\]

\[
8{,}\frac{4}{15} = \frac{124}{15}
\]

\[
4{,}\frac{2}{3} = \frac{14}{3} = \frac{70}{15}
\]

\[
\frac{124}{15} — \frac{70}{15} = \frac{54}{15}
\]

\[
5{,}\frac{2}{5} = \frac{27}{5}
\]

Делим обе части на \(\frac{27}{5}\):
\[
x = \frac{54}{15} \div \frac{27}{5} = \frac{54}{15} \cdot \frac{5}{27}
\]

\[
x = \frac{270}{405}
\]

Сокращаем на 135:
\[
x = \frac{2}{3}
\]

4)
\[
\frac{11}{24}x + 7{,}\frac{1}{3} = 9{,}\frac{5}{8}
\]

Переносим \(7{,}\frac{1}{3}\) вправо:
\[
\frac{11}{24}x = 9{,}\frac{5}{8} — 7{,}\frac{1}{3}
\]

\[
9{,}\frac{5}{8} = \frac{77}{8}
\]

\[
7{,}\frac{1}{3} = \frac{22}{3} = \frac{176}{24}
\]

\[
\frac{77}{8} = \frac{231}{24}
\]

\[
\frac{231}{24} — \frac{176}{24} = \frac{55}{24}
\]

Делим обе части на \(\frac{11}{24}\):
\[
x = \frac{55}{24} \div \frac{11}{24} = \frac{55}{24} \cdot \frac{24}{11}
\]

\[
x = 5
\]