ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 707

\[\boxed{\text{707.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} 2x^{2} + y^{2} = 9 \\ x^{2} - y^{2} = 3\ \ \\ \end{matrix} \right.\ ( + )\]

\[3x^{2} = 12\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[y^{2} = x^{2} - 3\]

\[y = \sqrt{x^{2} - 3}\]

\[x = 2:\]

\[y = \sqrt{4 - 3} = 1.\]

\[x = - 2\]

\[y = \sqrt{4 - 3} = 1.\]

\[Ответ:( - 2;1);\ \ (2;1).\]

\[\textbf{б)}\ \left\{ \begin{matrix} 2x^{2} - xy = 33 \\ 4x - y = 17\ \ \ \ \ \\ \end{matrix} \right.\ \ \]

\[\text{\ \ }\left\{ \begin{matrix} y = 4x - 17\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - x(4x - 17) = 0 \\ \end{matrix} \right.\ \ \]

\[\text{\ \ }\left\{ \begin{matrix} y = 4x - 17\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - 4x^{2} + 17x - 33 = 0 \\ \end{matrix} \right.\ \ \]

\[- 2x^{2} + 17x - 33 = 0\]

\[2x^{2} - 17x + 33 = 0\]

\[D = 289 - 8 \cdot 33 = 25\]

\[x_{1} = \frac{17 + 5}{4} = \frac{22}{4} = 5,5;\]

\[x_{2} = \frac{17 - 5}{4} = 3.\]

\[1)\ x_{1} = 3;\ \ \ y_{1} = - 5;\]

\[2)\ x_{2} = 5,5;\ \ y_{2} = 5.\]

\[Ответ:(3; - 5);(5,5;5).\]

\[\textbf{в)}\ \left\{ \begin{matrix} 3x^{2} - 2y = 1\ \ | \cdot 2 \\ 2x^{2} - y^{2} = 1\ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} 6x^{2} - 4y = 2\ \ \\ 6x^{2} - 3y^{2} = 3 \\ \end{matrix} \right.\ ( - )\]

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

\[3y^{2} - 4y + 1 = 0\]

\[D_{1} = 4 - 3 = 1\]

\[y_{1} = \frac{2 + 1}{3} = 1;\ \ y_{2} = \frac{2 - 1}{3} = \frac{1}{3}.\]

\[y = 1:\]

\[3x^{2} = 1 + 2 \cdot 1\]

\[3x^{2} = 3\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[y = \frac{1}{3}:\]

\[3x^{2} = 1 + 2 \cdot \frac{1}{3}\]

\[3x^{2} = \frac{5}{3}\]

\[x^{2} = \frac{5}{9}\]

\[x = \pm \frac{\sqrt{5}}{3}.\]

\[Ответ:( - 1;1);(1;1);\]

\[\left( - \frac{\sqrt{5}}{3};\frac{1}{3} \right);\ \ \left( \frac{\sqrt{5}}{3};\frac{1}{3} \right).\]

\[\textbf{г)}\ \left\{ \begin{matrix} x - y - 4 = 0 \\ x^{2} + y^{2} = 8,5\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x = y + 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 8y + 16 + y^{2} = 8,5 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} x = y + 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2y² + 8y + 7,5 = 0 \\ \end{matrix} \right.\ \]

\[2y^{2} + 8y + 7,5 = 0\]

\[D_{1} = 16 - 7,5 \cdot 2 = 1\]

\[y_{1} = \frac{- 4 + 1}{2} = - 1,5;\ \ \ \]

\[y_{2} = \frac{- 4 - 1}{2} = - 2,5.\]

\[1)\ y_{1} = - 1,5;\ \ \ \ x_{1} = 2,5;\]

\[2)\ y_{2} = - 2,5;\ \ \ \ x_{2} = 1,5.\]

\[Ответ:(1,5; - 2,5);(2,5;\ - 1,5).\]

\[\textbf{д)}\ \left\{ \begin{matrix} x^{2} + 4y = 10 \\ x - 2y = - 5 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\ \left\{ \begin{matrix} x = 2y - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} - 20y + 25 + 4y = 10 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 2y - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y² - 16y + 15 = 0 \\ \end{matrix} \right.\ \]

\[4y^{2} - 16y + 15 = 0\]

\[D_{1} = 64 - 60 = 4\]

\[y_{1} = \frac{8 + 2}{4} = \frac{10}{4} = 2,5;\]

\[y_{2} = \frac{8 - 2}{4} = \frac{6}{4} = 1,5.\]

\[1)\ y_{1} = 1,5;\ \ \ \ x_{1} = - 2;\]

\[2)\ y_{2} = 2,5;\ \ \ \ x_{2} = 0.\]

\[Ответ:( - 2;1,5);\ \ (0;2,5).\]

\[\textbf{е)}\ \left\{ \begin{matrix} x - 2y + 1 = 0 \\ 5xy + y^{2} = 16\ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = 2y - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5y(2y - 1) + y^{2} = 16 \\ \end{matrix} \right.\ \ \]

\[\text{\ \ }\left\{ \begin{matrix} x = 2y - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 11y² - 5y - 16 = 0 \\ \end{matrix} \right.\ \]

\[11y^{2} - 5y - 16 = 0\]

\[D = 25 + 11 \cdot 16 \cdot 4 = 729\]

\[y_{1} = \frac{5 + 27}{22} = \frac{32}{22} = \frac{16}{11} = 1\frac{5}{11};\]

\[y_{2} = \frac{5 - 27}{22} = - 1.\]

\[1)\ y_{1} = - 1\ \ \ \ \ \ \ \ x_{1} = - 3;\]

\[2)\ y_{2} = 1\frac{5}{11};\ \ x_{2} = 1\frac{10}{11}.\]

\[Ответ:( - 3; - 1);\ \ \left( 1\frac{10}{11};1\frac{5}{11} \right).\]