Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. Задачи на повторение 36

\[\boxed{\mathbf{П.\ 36}}\]

\[\textbf{а)}\ \frac{3}{7}x + 5\frac{1}{2}x - 7\frac{3}{4} =\]

\[= 1 - \frac{4}{7}x + 5\frac{1}{4}x\]

\[\frac{3}{7}x + \frac{4}{7}x + 5\frac{2}{4}x - 5\frac{1}{4}x =\]

\[= 1 + 7\frac{3}{4}\]

\[\frac{7}{7}x + \frac{1}{4}x = 8\frac{3}{4}\]

\[x + \frac{1}{4}x = 8\frac{3}{4}\]

\[\frac{5}{4}x = \frac{35}{4}\]

\[x = \frac{35}{4}\ :\frac{5}{4} = \frac{35}{4} \cdot \frac{4}{5}\]

\[x = 7.\]

\[\textbf{б)}\ 5 - 2\frac{1}{3}z + 4\frac{4}{9}z =\]

\[= 7\frac{1}{2}z - 6\frac{5}{12}z + 6\frac{1}{3}\]

\[- 2\frac{3}{9}z + 4\frac{4}{9}z - 7\frac{6}{12}z + 6\frac{5}{12}z =\]

\[= 6\frac{1}{3} - 5\]

\[2\frac{1}{9}z - z = 1\frac{1}{3}\]

\[1\frac{1}{9}z = 1\frac{1}{3}\]

\[z = \frac{4}{3}\ :\frac{10}{9} = \frac{4}{3} \cdot \frac{9}{10}\]

\[z = \frac{12}{10} = 1,2.\]

\[\textbf{в)}\ 4 \cdot \left( \frac{2}{7}n + 1 \right) + 2\frac{1}{2} =\]

\[= 6 - \frac{1}{3} \cdot \left( \frac{6}{7}n - 3 \right)\]

\[\frac{8}{7}n + 4 + 2\frac{1}{2} = 6 - \frac{2}{7}n + 1\]

\[\frac{8}{7}n + \frac{2}{7}n = 7 - 6\frac{1}{2}\]

\[\frac{10}{7}n = \frac{1}{2}\]

\[n = \frac{1}{2}\ :\frac{10}{7} = \frac{1}{2} \cdot \frac{7}{10}\]

\[n = \frac{7}{20} = 0,35.\]

\[\textbf{г)}\ 2 - \left( 1\frac{1}{3}p + \frac{1}{7} \right) \cdot 21 =\]

\[= 4\frac{1}{4}p - 6\frac{3}{8}\]

\[2 - \frac{4}{3}p \cdot 21 - 3 = 4\frac{1}{4}p - 6\frac{3}{8}\]

\[2 - 28p - 3 = 4\frac{1}{4}p - 6\frac{3}{8}\]

\[- 28p - 4\frac{1}{4}p = - 6\frac{3}{8} + 1\]

\[- 32\frac{1}{4}p = - 5\frac{3}{8}\]

\[- \frac{129}{4}p = - \frac{43}{8}\]

\[p = \frac{43}{8}\ :\frac{129}{4} = \frac{43}{8} \cdot \frac{4}{129}\]

\[p = \frac{1}{6}.\]