Решебник по математике 6 класс Мерзляк ФГОС Задание 363

 

\[\boxed{\mathbf{363\ (363).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

Пояснение.

Решение.

\[= 9x + \frac{26}{3}y - \frac{1}{2}z =\]

\[= 9x + 8\frac{2}{3}y - \frac{1}{2}z\]

\[\boxed{\mathbf{363}\mathbf{.}}\]

\[1)\ \frac{4^{\backslash 2}}{9} + \frac{1^{\backslash 3}}{6} = \frac{8 + 3}{18} = \frac{11}{18}\ (У)\]

\[2)\ 2 - 1\frac{5}{18} = 1\frac{18}{18} - 1\frac{5}{18} =\]

\[= \frac{13}{18}\ (Л)\]

\[3)\ \frac{22}{45} \cdot \frac{35}{44} = \frac{1 \cdot 7}{9 \cdot 2} = \frac{7}{18}\ (А)\]

\[4)\ 1\frac{7}{27}\ :4\frac{8}{15} = \frac{34}{27}\ :\frac{68}{15} =\]

\[= \frac{34 \cdot 15}{27 \cdot 68} = \frac{1 \cdot 5}{9 \cdot 2} = \frac{5}{18}\ (Н)\]

\[5)\ 2\frac{4^{\backslash 2}}{9} - 1\frac{1^{\backslash 9}}{2} = 2\frac{8}{18} - 1\frac{9}{18} =\]

\[= 1\frac{26}{18} - 1\frac{9}{18} = \frac{17}{18}\ (О)\]

\[6)\ \frac{2^{\backslash 6}}{3} + \frac{7}{18} = \frac{12}{18} + \frac{7}{18} = \frac{19}{18} =\]

\[= 1\frac{1}{18}\ (В)\]

\[7)\ 2\frac{5}{9}\ :2 = \frac{23}{9} \cdot \frac{1}{2} = \frac{23}{18} =\]

\[= 1\frac{5}{18}\ (А)\ \]

\[Балерина:\]

\[УЛАНОВА.\]