Решебник по геометрии 9 класс Атанасян ФГОС Задание 1068

 

\[\boxed{\mathbf{1068.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\]

\[\left| \overrightarrow{a} \right| = 2;\]

\[\left| \overrightarrow{b} \right| = 5;\]

\[\widehat{\overrightarrow{a}\overrightarrow{b}} = 120{^\circ};\]

\[\overrightarrow{p} = x\overrightarrow{a} + 17\overrightarrow{b};\]

\[\overrightarrow{q} = 3\overrightarrow{a} - \overrightarrow{b};\]

\[\overrightarrow{p}\bot\overrightarrow{q}.\]

\[\mathbf{Найти:}\]

\[x - ?\]

\[\mathbf{Решение.}\]

\[1)\ \overrightarrow{p} \bullet \overrightarrow{q} =\]

\[= \left( x\overrightarrow{a} + 17\overrightarrow{b} \right)\left( 3\overrightarrow{a} - \overrightarrow{b} \right) =\]

\[= 3x{\overrightarrow{a}}^{2} - x\overrightarrow{a}\overrightarrow{b} + 51\overrightarrow{a}\overrightarrow{b} - 17{\overrightarrow{b}}^{2} =\]

\[= 12x + 5x - 225 - 425 =\]

\[= 17x - 680.\]

\[2)\ \overrightarrow{p}\bot\overrightarrow{q} \Longrightarrow \overrightarrow{p} \bullet \overrightarrow{q} = 0:\]

\[17x - 680 = 0\]

\[x = \frac{680}{17} = 40.\]

\[Ответ:\ x = 40.\]

\[\boxed{\mathbf{1068.еуроки - ответы\ на\ пятёрку}}\]

\[\textbf{а)}\ (x + 2)^{2} + y^{2} = 1;\]

\[x^{2} + y^{2} = 4.\]

\[\left\{ \begin{matrix} (x + 2)^{2} + y^{2} = 1 \\ x^{2} + y^{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ ( - )\]

\[(x + 2)^{2} - x^{2} = - 3\]

\[x^{2} + 4x + 4 - x^{2} = - 3\]

\[4x = - 7\]

\[x = - \frac{7}{4}.\]

\[y^{2} = 4 - x^{2} = 4 - \left( \frac{7}{4} \right)^{2} =\]

\[= 4 - \frac{49}{16} = \frac{64 - 49}{16} = \frac{15}{16}\]

\[y = \pm \frac{\sqrt{15}}{4}.\]

\[Ответ:две\ точки\ пересечения.\]

\[\textbf{б)}\ (x + 3)^{2} + y^{2} = 1;\]

\[x^{2} + y^{2} = 4.\ \]

\[\left\{ \begin{matrix} (x + 3)^{2} + y^{2} = 1 \\ x^{2} + y^{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ ( - )\]

\[(x + 3)^{2} - x^{2} = - 3\]

\[x^{2} + 6x + 9 - x^{2} = - 3\]

\[6x = - 12\]

\[x = - 2.\]

\[y^{2} = 4 - x^{2} = 4 - 4 = 0;\]

\[y = 0.\]

\[Ответ:одна\ точка\ \]

\[пересечения\ (касания).\]