\[\boxed{\mathbf{960.ОК\ ГДЗ - домашка\ на}\ 5}\]
\(Дано:\)
\[A(3; - 4);B(1;0);\]
\[C(0;5);D(0;0);E(0;1).\]
\[\textbf{а)}\ x^{2} + y^{2} = 25\]
\[A(3; - 4)\ :\ \ \ \ \]
\[9 + 16 = 25 - верно.\]
\[C(0;5)\ :\ \ \ \ \]
\[0 + 25 = 25 - верно.\]
\[\textbf{б)}\ (x - 1)^{2} + (y + 3)^{2} = 9\]
\[B(1;0)\ :\ \ \ \]
\[0 + 9 = 9 - верно.\]
\[\textbf{в)}\ \left( x - \frac{1}{2} \right)^{2} + y^{2} = \frac{1}{4}\]
\[B(1;0)\ :\ \ \ \ \]
\[\left( \frac{1}{2} \right)^{2} + 0 = \frac{1}{4} - верно.\]
\[D(0;0)\ :\ \ \ \ \]
\[\left( - \frac{1}{2} \right)^{2} + 0 = \frac{1}{4} - верно.\]
\[Ответ:а)\ A\ и\ C;б)\ B;в)\ \text{B\ }и\ \text{D.}\]
\[\boxed{\mathbf{960.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм.\]
\[\mathbf{Доказать:}\]
\[\overrightarrow{\text{XA}} + \overrightarrow{\text{XC}} = \overrightarrow{\text{XB}} + \overrightarrow{\text{XD}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ \overrightarrow{\text{XA}} - \overrightarrow{\text{XB}} = \overrightarrow{\text{BX}} + \overrightarrow{\text{XA}} = \overrightarrow{\text{BA}};\]
\[\overrightarrow{\text{XD}} - \overrightarrow{\text{XC}} = \overrightarrow{\text{CX}} + \overrightarrow{\text{XD}} =\]
\[= \overrightarrow{\text{CD}}\text{\ \ }(по\ правилу\ треугольника).\]
\[2)\ ABCD - параллелограмм:\]
\[\overrightarrow{\text{BA}} \uparrow \uparrow \overrightarrow{\text{CD}}\text{\ \ }и\ \left| \overrightarrow{\text{BA}} \right| = \left| \overrightarrow{\text{CD}} \right|.\]
\[Следовательно:\ \]
\[\overrightarrow{\text{BA}} = \overrightarrow{\text{CD}}.\]
\[3)\ Получаем:\ \]
\[\overrightarrow{\text{XA}} - \overrightarrow{\text{XB}} = \overrightarrow{\text{XD}} - \overrightarrow{\text{XC}};\]
\[\overrightarrow{\text{XA}} + \overrightarrow{\text{XC}} = \overrightarrow{\text{XB}} + \overrightarrow{\text{XD}}.\]
\[Что\ и\ требовалось\ доказать.\]