Решебник по математике 6 класс Мерзляк ФГОС Задание 362

 

\[\boxed{\mathbf{362\ (362).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 0,5x + 1,4 - \frac{7}{18}x - \frac{1}{9}x =\]

\[= \left( \frac{1^{\backslash 9}}{2} - \frac{7}{18} - \frac{1^{\backslash 2}}{9} \right)x + 1,4 =\]

\[= \frac{9 - 7 - 12}{18}x + 1,4 =\]

\[= \frac{9 - 9}{18}x + 1,4 = 0 \cdot x + 1,4 =\]

\[= 1,4\]

\[2)\ 0,5x + \frac{1}{7}x + 7,4 - \frac{9}{14}x =\]

\[= \left( \frac{1^{\backslash 7}}{2} + \frac{1^{\backslash 2}}{7} - \frac{9}{14} \right)x + 7,4 =\]

\[= \frac{7 + 2 - 9}{14}x + 7,4 =\]

\[= \frac{9 - 9}{14}x + 7,4 = 0 \cdot x + 7,4 =\]

\[= 7,4\]

\[3)\ 1\frac{17}{18}a + 1 - 1,5a - \frac{4}{9}a =\]

\[= \left( \frac{35}{18} - \frac{3^{\backslash 9}}{2} - \frac{4^{\backslash 2}}{9} \right)a + 1 =\]

\[= \frac{35 - 27 - 8}{18}a + 1 =\]

\[= \frac{35 - 35}{18}a + 1 = 0 \cdot a + 1 = 1\]

\[4)\ 2,4 + 1,25b + \frac{5}{6}b - 2\frac{1}{12}b =\]

\[= \left( \frac{5^{\backslash 3}}{4} + \frac{5^{\backslash 4}}{6} - \frac{25}{12} \right)b + 2,4 =\]

\[= \frac{15 + 10 - 25}{12}b + 2,4 =\]

\[= \frac{25 - 25}{12}b + 2,4 =\]

\[= 0 \cdot b + 2,4 = 2,4\]

\[\boxed{\mathbf{362}\mathbf{.}}\]

\[1)\ \left( \frac{5^{\backslash 2}}{12} + \frac{1^{\backslash 3}}{8} \right)\ :\frac{3}{8} =\]

\[= \frac{10 + 3}{24} \cdot \frac{8}{3} = \frac{13 \cdot 8}{24 \cdot 3} = \frac{13}{9} = 1\frac{4}{9}\]

\[2)\ \frac{5}{12} + \frac{1}{8}\ :\frac{3}{8} = \frac{5}{12} + \frac{1}{8} \cdot \frac{8}{3} =\]

\[= \frac{5}{12} + \frac{1^{\backslash 4}}{3} = \frac{5 + 4}{12} = \frac{9}{12} = \frac{3}{4}\]