ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 50

 

\[\boxed{\text{50\ (50).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x = 16\ :( - 5)\]

\[x = \frac{16}{- 5}\]

\[x = - 3,2.\]

\[x = \frac{1}{5}\ :2\]

\[x = \frac{1}{10}\]

\[x = 0,1.\]

\[x = 4 \cdot 3\]

\[x = 12.\]

\[\textbf{г)}\ 4x = - 2\]

\[x = - 2\ :4\]

\[x = - \frac{2}{4}\]

\[x = - 0,5.\]

\[x = 3\ :0,6\]

\[x = 5.\]

\[x = 5\ :( - 0,7)\]

\[x = 5\ :\left( - \frac{7}{10} \right)\]

\[x = - 5\ \cdot \frac{10}{7}\]

\[x = - \frac{50}{7}\]

\[x = - 7\frac{1}{7}\text{.\ }\]

\[\boxed{\text{50.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 2a + b = \frac{2a + b}{1} =\]

\[= \frac{(2a + b) \cdot b}{1 \cdot b} = \frac{2ab + b^{2}}{b}\]

\[\textbf{б)}\ 2a + b = \frac{2a + b}{1} =\]

\[= \frac{(2a + b) \cdot 5}{1 \cdot 5} = \frac{10a + 5b}{5}\]

\[\textbf{в)}\ 2a + b = \frac{2a + b}{1} =\]

\[= \frac{(2a + b) \cdot 3a}{1 \cdot 3a} = \frac{6a^{2} + 3ab}{3a}\]

\[\textbf{г)}\ 2a + b = \frac{2a + b}{1} =\]

\[= \frac{(2a + b) \cdot (2a - b)}{1 \cdot (2a - b)} =\]

\[= \frac{4a^{2} - b^{2}}{2a - b}\text{\ \ }\]

\[\text{\ \ }\]