ГДЗ по геометрии 9 класс Атанасян Задание 781

\[\boxed{\mathbf{781.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\]

\[\overrightarrow{x} = \overrightarrow{m} + \overrightarrow{n};\]

\[\overrightarrow{y} = \overrightarrow{m} - \overrightarrow{n}\text{.\ \ \ \ \ }\]

\[Выразить:\]

\[\textbf{а)}\ 2\overrightarrow{x} - 2\overrightarrow{y};\]

\[\textbf{б)}\ 2\overrightarrow{x} + \frac{1}{2}\overrightarrow{y};\]

\[\textbf{в)} - \overrightarrow{x} - \frac{1}{3}\overrightarrow{y}\mathbf{.}\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ 2\overrightarrow{x} - 2\overrightarrow{y} =\]

\[= 2 \bullet \left( \overrightarrow{m} + \overrightarrow{n} \right) - 2 \bullet \left( \overrightarrow{m} - \overrightarrow{n} \right) =\]

\[= 2\overrightarrow{m} + 2\overrightarrow{n} - 2\overrightarrow{m} + 2\overrightarrow{n} = 4\overrightarrow{n}.\]

\[\textbf{б)}\ 2\overrightarrow{x} + \frac{1}{2}\overrightarrow{y} =\]

\[= 2 \bullet \left( \overrightarrow{m} + \overrightarrow{n} \right) + \frac{1}{2} \bullet \left( \overrightarrow{m} - \overrightarrow{n} \right) =\]

\[= 2\overrightarrow{m} + 2\overrightarrow{n} + \frac{1}{2}\overrightarrow{m} - \frac{1}{2}\overrightarrow{n} =\]

\[= 2,5\overrightarrow{m} + 1,5\overrightarrow{n}.\]

\[\textbf{в)}\ - \overrightarrow{x} - \frac{1}{3}\overrightarrow{y} =\]

\[= - \left( \overrightarrow{m} + \overrightarrow{n} \right) - \frac{1}{3} \bullet \left( \overrightarrow{m} + \overrightarrow{n} \right) =\]

\[= - \overrightarrow{m} - \overrightarrow{n} - \frac{1}{3}\ \overrightarrow{m} + \frac{1}{3}\overrightarrow{n} =\]

\[= - 1\frac{1}{3}\overrightarrow{m} - \frac{2}{3}\overrightarrow{n}.\]