\[\boxed{\mathbf{988.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Дано:\ \]
\[\overrightarrow{a}\ и\ \overrightarrow{b}\ не\ коллинеарны.\]
\[Найти:\]
\[x,\ чтобы\ \overrightarrow{p}\ и\ \overrightarrow{q} - коллинеарны.\]
\[Решение.\]
\[\textbf{а)}\ \overrightarrow{p} = 2\overrightarrow{a} - \overrightarrow{b};\ \ \ \overrightarrow{q} = \overrightarrow{a} + x\overrightarrow{b}:\]
\[\frac{2}{1} = - \frac{1}{x}\]
\[2x = - 1\]
\[x = - \frac{1}{2}.\]
\[\textbf{б)}\ \overrightarrow{p} = x\overrightarrow{a} - \overrightarrow{b};\ \ \ \overrightarrow{q} = \overrightarrow{a} + x\overrightarrow{b}:\]
\[\frac{x}{1} = - \frac{1}{x}\]
\[x^{2} = - 1 - нет\ решений;\]
\[\ \overrightarrow{p}\ и\ \overrightarrow{q}\ не\ коллинеарны.\]
\[\textbf{в)}\ \overrightarrow{p} = \overrightarrow{a} + x\overrightarrow{b};\ \ \ \overrightarrow{q} = \overrightarrow{a} - 2\overrightarrow{b}:\]
\[\frac{1}{1} = - \frac{x}{2}\]
\[x = - 2.\]
\[\textbf{г)}\ \overrightarrow{p} = 2\overrightarrow{a} + \overrightarrow{b};\ \ \ \overrightarrow{q} = \overrightarrow{\text{xa}} + \overrightarrow{b}:\]
\[\frac{2}{x} = \frac{1}{1}\]
\[2 = x\]
\[x = 2.\]
\[\boxed{\mathbf{988}\mathbf{.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\textbf{а)}\ \overrightarrow{m} \nearrow \nearrow \overrightarrow{n};\ \]
\[\textbf{б)}\ \overrightarrow{m} \nearrow \nearrow \overrightarrow{n}\ и\ \ \left| \overrightarrow{m} \right| \geq \overrightarrow{n}.\]
\[\mathbf{Доказать:}\]
\[\textbf{а)}\ \left| \overrightarrow{m} + \overrightarrow{n} \right| = \left| \overrightarrow{m} \right| + \left| \overrightarrow{n} \right|;\]
\[\textbf{б)}\ \left| \overrightarrow{m} + \overrightarrow{n} \right| = \left| \overrightarrow{m} \right| - \left| \overrightarrow{n} \right|.\]
\[\mathbf{Доказательство.}\]
\[\textbf{а)}\ \left| \overrightarrow{\text{AB}} \right| + \left| \overrightarrow{\text{BC}} \right| = \left| \overrightarrow{\text{AC}} \right|:\]
\[\ \left| \overrightarrow{m} + \overrightarrow{n} \right| = \left| \overrightarrow{m} \right| + \left| \overrightarrow{n} \right|.\]
\[\textbf{б)}\ AB = AC + BC:\]
\[\overrightarrow{\text{AC}} = \left| \overrightarrow{m} \right| + \left| \overrightarrow{n} \right|\ \]
\[\left| \overrightarrow{\text{AC}} \right| = \left| \overrightarrow{\text{AB}} \right| - \left| \overrightarrow{\text{BC}} \right|.\]
\[Следовательно:\]
\[\left| \overrightarrow{m} \right| + \left| \overrightarrow{n} \right| = \left| \overrightarrow{m} \right| - \left| \overrightarrow{n} \right|\]