\[\boxed{\mathbf{1068.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Дано:}\]
\[\left| \overrightarrow{a} \right| = 2;\]
\[\left| \overrightarrow{b} \right| = 5;\]
\[\widehat{\overrightarrow{a}\overrightarrow{b}} = 120{^\circ};\]
\[\overrightarrow{p} = x\overrightarrow{a} + 17\overrightarrow{b};\]
\[\overrightarrow{q} = 3\overrightarrow{a} - \overrightarrow{b};\]
\[\overrightarrow{p}\bot\overrightarrow{q}.\]
\[\mathbf{Найти:}\]
\[x - ?\]
\[\mathbf{Решение.}\]
\[1)\ \overrightarrow{p} \bullet \overrightarrow{q} =\]
\[= \left( x\overrightarrow{a} + 17\overrightarrow{b} \right)\left( 3\overrightarrow{a} - \overrightarrow{b} \right) =\]
\[= 3x{\overrightarrow{a}}^{2} - x\overrightarrow{a}\overrightarrow{b} + 51\overrightarrow{a}\overrightarrow{b} - 17{\overrightarrow{b}}^{2} =\]
\[= 12x + 5x - 225 - 425 =\]
\[= 17x - 680.\]
\[2)\ \overrightarrow{p}\bot\overrightarrow{q} \Longrightarrow \overrightarrow{p} \bullet \overrightarrow{q} = 0:\]
\[17x - 680 = 0\]
\[x = \frac{680}{17} = 40.\]
\[Ответ:\ x = 40.\]
\[\boxed{\mathbf{1068.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ (x + 2)^{2} + y^{2} = 1;\]
\[x^{2} + y^{2} = 4.\]
\[\left\{ \begin{matrix} (x + 2)^{2} + y^{2} = 1 \\ x^{2} + y^{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ ( - )\]
\[(x + 2)^{2} - x^{2} = - 3\]
\[x^{2} + 4x + 4 - x^{2} = - 3\]
\[4x = - 7\]
\[x = - \frac{7}{4}.\]
\[y^{2} = 4 - x^{2} = 4 - \left( \frac{7}{4} \right)^{2} =\]
\[= 4 - \frac{49}{16} = \frac{64 - 49}{16} = \frac{15}{16}\]
\[y = \pm \frac{\sqrt{15}}{4}.\]
\[Ответ:две\ точки\ пересечения.\]
\[\textbf{б)}\ (x + 3)^{2} + y^{2} = 1;\]
\[x^{2} + y^{2} = 4.\ \]
\[\left\{ \begin{matrix} (x + 3)^{2} + y^{2} = 1 \\ x^{2} + y^{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ ( - )\]
\[(x + 3)^{2} - x^{2} = - 3\]
\[x^{2} + 6x + 9 - x^{2} = - 3\]
\[6x = - 12\]
\[x = - 2.\]
\[y^{2} = 4 - x^{2} = 4 - 4 = 0;\]
\[y = 0.\]
\[Ответ:одна\ точка\ \]
\[пересечения\ (касания).\]